
























Class J 2 \£j.\ 53, 

Book._A_ 





Copyright N'.’_ \3S, 8 


COPYRIGHT DEPOSIT. 


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NEW 

ELEMENTARY ALGEBRA 


FOR NEW YORK 
REGENTS’ SCHOOLS 

BY 

WEBSTER WELLS, S.B. 

n 7 

AUTHOR OF A SERIES OF TEXTS ON MATHEMATICS 

AND 

WALTER W. HART, A.B. 

ASSOCIATE PROFESSOR OF MATHEMATICS, UNIVERSITY OF WISCONSIN 

SCHOOL OF EDUCATION 



D. C. HEATH AND COMPANY 

BOSTON NEW YORK CHICAGO 



Q A i5& 

i <{ a ^ 


COPYRIGHT, 1923, 1924 AND 1928 
BY D. C. HEATH & CO. 

2 E 8 


This book may be had 
with answers or without answers 
at the same price. 

Answer books, bound in paper, may be 
obtained free of charge by teachers . 



©Cl A1077304 


PRINTED IN U. S. A. 

MAY 19 1928 



PREFACE 


This text meets the present requirements (1928) in Elementary- 
Algebra set by the Regents of the University of the State of 
New York. It is based upon a text produced in 1924, in the 
preparation of which suggestions, criticisms, and actual contri¬ 
butions of material were sought and received from a number 
of experienced teachers of mathematics of the State of New 
York, of whom Messrs. G. R. Bodley of Fulton, Elmer Schuyler 
of Bay Ridge High School (Brooklyn), H. J. Lathrop of the 
Brockport Normal School, and L. J. Hellriegel of the Masten 
Park High School (Buffalo) merit special mention. In the 
preparation of this text, the author and publishers acknowledge 
special indebtedness to Mr. Schuyler. 

Contents of the text. It is sufficient to state that the topics 
required in Elementary Algebra by the Regents of the University 
of the State of New York comprise the major part of this text. 
All topics which are marked by a star in the Syllabus to indicate 
that they are optional, are also marked by a star in this text. 
(See pages 120, 122.) 

There appear in the text a few topics which are required by 
the College Entrance Examination Board (C.E.E.B.) and a 
little additional material on the quadratic, which are not in¬ 
cluded even as optional material in the Syllabus; all this material 
is plainly marked. (See pages 122, 128, 239, 250.) 

The additional claims of merit for this text are the arrange¬ 
ment of the subject matter, the quality of the instruction, in 
general, and the many particular devices designed to help the 
teacher develop in the pupils mastery of algebra. In these 
respects, this text has in enlarged degree the qualities which 
have made the Wells and Hart Algebras notable for over fifteen 
years, — in “ enlarged degree,” because its author has been 

iii 





IV 


PREFACE 


teaching algebra to boys and girls of high school age during 
this time, as well as writing texts for them. 

The keynote of the arrangement of topics and of the instruc¬ 
tion is the practice of teaching one idea at a time, doing it well, 
and then repeating the topic until it has been mastered. 

1. The emphasis upon small units of instruction is illustrated 
by the separate instruction on the use of the axioms by which 
equations are solved (pages 17, 43, 47, 78); on the gradual 
extension of the use of formulas (see the first stage, pages 3-6; 
the second stage, pages 21-22; the separate instruction on each 
of the more difficult types of problems (see pages 80, 82, 102, 
176, etc.). 

2. The thoroughness of the instruction is illustrated on such 
pages as 33, 48, 112, etc. The explanations are fully expressed 
in simple language, and are invariably inductive in style. They 
have been described as “ the kind that experienced teachers use 
in the classroom.” 

3. The following special devices should be noted. 

a. Each topic taught is used, if possible, in the solution of 
equations or of formulas. 

This gives a motive for the study of the topic, permits good gradation 
of the material, and properly emphasizes the equation and the formula. 

b. Translation of numerical relations into algebraic symbols, 
and the solution of problems receives unique treatment. 

Each of the more difficult types is taught separately; the types are 
introduced gradually, according to their difficulty; and all the types are 
repeated after their first appearance, in miscellaneous lists of problems. 
Observe the use of “charts” as an aid in solving problems. (See pages 
80, 82, 83, 100, etc.) 

c. Factoring and special products are taught together, — the 
natural way to study and learn them. 

d. Thoughtful solution of equations is compelled by subordinat¬ 
ing “ transposition ” and “ clearing of fractions.” 

Observe the omission of these processes until page 96. Observe the 
symbols A, S, M, and D, which are used in explaining the solution of 
equations. (See pages 17, 43, 47, and 78.) 


PREFACE 


V 


c. Graphs of statistics, and line graphs are introduced early in 
order that they may contribute to growth of the concept of 
functionality. The subject is then extended in Chapter XI, 
just before simultaneous equations. 

/. Abundance of examples and problems. 

Teachers will not need to waste time dictating examples, because 
enough examples appear in the text, and they are the kind that pupils 
can solve with profit. Observe that those in the body of the text are 
supplemented by other classified lists at the end of the text, which are 
referred to opportunely by such notes as those on pages 77, 93, etc. 

If there are more examples than are needed for a class, as may happen 
in some cases, teachers are advised to teach the odd examples one year 
and the even ones the next year, — thus obtaining a selection of well- 
graded examples for each class. 

g. Suitable reviews appear periodically and also at the end of 
the text. (See pages 39, 47, 60, 71, etc.) 

One reader has said, “These reviews should prove invaluable.” 

h. Concerning oral algebra. 

No examples are marked “for oral solution.” Every teacher should 
and will have pupils do orally all easy examples which are encountered 
during class work, and, also, will have pupils write only the answers 
to other easy examples which may be assigned for the study period. 
This is a matter of judgment, which is properly left to the individual 
teacher. In every list of examples in this text, there will be found at 
the beginning a number of examples which can be done orally, — except 
in such obvious cases as multiplication, division, and square root of 
polynomials, etc. Teachers are urged to have many such done orally 
in class. 


April, 1928 


CONTENTS 


CHAPTER PAGE 

I. Literal Numbers — Formulas.1 

II. Positive and Negative Numbers .... 33 

Addition of Positive and Negative Numbers . . 36 

Multiplication of Positive and Negative Numbers . 38 

III. Addition and Subtraction of Polynomials . . 41 

Addition of Monomials ...... 42 

Addition of Polynomials.45 

Subtraction ......... 50 

Subtraction of Positive and Negative Numbers . . 51 

Subtraction of Polynomials.52 

IV. Parentheses.61 

Removal of Parentheses.61 

Introduction of Parentheses.64 

V. Multiplication.72 

Multiplication of Monomials.73 

Multiplication of Polynomials by Monomials . . 74 

Multiplication of a Polynomial by a Polynomial . . 76 

VI. Division.86 

Division of a Monomial by a Monomial ... 88 

Division of Polynomials by Monomials ... 90 

Division of Polynomials by Polynomials ... 92 

VII. Simple Equations.95 

VIII. Special Products and Factoring .... 107 

Quadratic Equations by Factoring . . . .134 

IX. Fractions.139 

Reduction of Fractions.140 

Multiplication of Fractions.142 

Division of Fractions.145 

Addition and Subtraction of Fractions . . . .153 


vi 










CONTENTS 


vii 

CHAPTER PAGE 

X. Fractional Equations.168 

Ratio and Proportion.183 

XI. Graphical Representation.188 

XII. Linear Equations Having Two Unknowns . .201 

Literal Simultaneous Equations.220 

XIII. Square Root and Radicals. 224 

Quadratic Surds.233 

XIV. Quadratic Equations. 242 

Complete Quadratic Equations.245 

XV. Trigonometry of the Right Triangle ... 258 

XVI. Variation. 272 

Supplementary Exercises.278 

Miscellaneous Reviews.305 

Regents’ Examinations.316 

Tables .331 

Index.332 








Special Notes to the Teacher: 

I. Concerning Accuracy in Measuring and Computing. 

For an extended discussion of this subject, see the Tentative Syllabus 
in Junior High School Mathematics, 1927, pages 49 to 54. 

Measures are necessarily only approximate. Thus the radius 
of a circle, given as 3.5, may be as much as 3.54 or as little as 
3.46. All that can be said of it is that it is 3.5, correct to tenths 
or correct to two significant figures. 

Numerical measures must often be “ rounded off.” One 
accepted practice is not well known. When a final digit 5 follows 
an even digit, that even digit is retained when the final digit 5 is 
dropped in the process of rounding off the original number to one 
fewer significant figures. (See above reference, p. 50.) 

Thus, 3.445, rounded off to three figures, becomes 3.44; whereas 
3.455, rounded off to three figures, becomes 3.46. 

Two plans for computing with numerical measures. 

a. Round off each of the numbers used to the number of sig¬ 
nificant figures which appears in the least accurate number; com¬ 
pute with the resulting numbers ; round off the final result to the 
same number of significant figures. (See above reference, p. 53.) 

b. Compute with the numbers as given; round off the final 
result to the number of significant figures which appears in the 
least accurate number. (See above reference, p. 53.) 

Apparently the latter plan is preferable at present. It will 
be followed when computing the answers for this text. Both the 
original and the “rounded off” result will be given. 

II. On the significance of the symbols * and # in this text. 

The symbol * is employed to designate sections and examples 
which are not required in the minimum course by the Syllabus 
in Elementary Algebra (1928), but which are suggested as desir¬ 
able optional topics. These sections are in some cases required by 
the College Entrance Examination Board. Explanation appears 
with these sections. 

The symbol # is employed to designate sections or examples 
which are not required by the Syllabus or even recommended as 
desirable optional topics. These, obviously, should not be studied 
unless the class has mastered the required and the optional topics of 
the Syllabus. They are included either because they are required 
by the College Entrance Examination Board, or because they are 
desirable extensions of the minimum requirements to use as chal¬ 
lenges for able sections or able pupils. 


ALGEBRA 


I. LITERAL NUMBERS - FORMULAS 

1. Letters are used to represent numbers in algebra. 

Thus, p = b X r 

expresses the rule 

the percentage equals the base multiplied by the rate. 

If b = 2500, and r = 5 %, 
p =- 2500 X .05, or 125. 

A letter used to represent a number is called a literal number. 

The use of literal numbers is one characteristic of algebra. 

2. Signs of multiplication. The symbol X is not used much 
in algebra as the sign of multiplication because it is like the 
letter x. A dot, *, placed above the line between two numbers, 
means that the numbers are to be multiplied together. 

Thus, 6 • 7 means 6 times 7, or 42. 

a • b means a times b. 

Even the dot is omitted when one or both of the numbers are 
literal numbers. 

Thus, 6 a means 6 times a. It is read “six a” 
xt means x times t. It is read “ex tee.” 

Historical Note. — The symbol X was first used by an English¬ 
man, Oughtred, about 1631. The symbol • was introduced by Leib¬ 
nitz in 1693. Multiplication was indicated in Hindu and Italian books 
by writing the factors side by side, as early as the thirteenth century. 

3. The product of two or more numbers is the result obtained 
by multiplying together the numbers; the numbers themselves 
are the factors of the product. Every factor of a product is an 
exact divisor of the product. 

Thus, 5 and 8 are factors of 40. 

1 


2 


ALGEBRA 


EXERCISE 1 

1. What does 3 t mean ? .45 to ? 2 .8 xy? r -s • £ ? 

2. How much is, or what is the value of: 

а. 8 y when y is 7, 10, 12, 5,-J-, £? 

б. 24 m when m is ■§■, f, f, f, -f-J ? 

c. 18 t when t is .5, .8, .03, -J, § ? 

3. a. What are the values of 9 y when y is 3, 5, 6, 10, 20 ? 

b. As y increases (becomes greater), what happens to 9 y ? 

4. a. What are the values of 10 M when M is 15, 12, 10, 9 ? 

b. As M decreases (becomes smaller), what happens to 10 M ? 

6. A certain number is represented by n. What repre¬ 
sents the number 4 times as large? 11 times as large? 

6. How many cents are there in 3 dollars ? In 5 dollars ? 
In a dollars ? In x quarters ? In m dimes ? 

7. a. How much do 5 doz. eggs cost at 40^ per dozen ? At 
50^ per dozen? At x cents per dozen? 

b. How much is 5 x cents when x is 60 ? 

8. a. How much do 8 tons of coal cost at $12 per ton ? 

b. How much do 8 tons of coal cost at p dollars per ton? 

9. An automobile traveled m miles an hour. How far did 
it travel in 6 hr. ? In 8 hr. ? In 30 min. ? 

10. A girl saved t cents each week. How much did she 
save in 18 weeks ? In 35 weeks ? 

11. An article costs $3. How much do n of them cost? j 

12. Represent the product of y and 7.5. 

13. a. An article cost $5. It was sold so that the profit was 
10 % of the cost. What was the profit ? 

b. An article cost C dollars. It was sold so that the profit 
was 10 % of the cost. What was the profit ? 

14. a. What is the interest on $100 for 1 yr. at 5 % ? 

b. What is the interest on P dollars for 1 yr. at 5 % ? 


LITERAL NUMBERS —FORMULAS 


3 


4. A formula is a rule of computation expressed by means 
of arithmetical and literal numbers, connected by mathematical 
signs which tell what must be done with the numbers. 

Thus, in § 1, p = br is a formula. It is the percentage formula. 

1. The formula for the area of a ^ ^ 

rectangle. You know the rule: - 

The area of a rectangle equals the prod- h 
uct of its base and altitude. 

Thus, if the base is 8 in., and the altitude E b F 

is 15 in., the area is 8 X 15 or 120 sq. in. 

If h = the number of units of length in the altitude, 
b = the number of the same units in the base, 

and A = the number of the corresponding surface units in 
the interior of the rectangle, 
then A = hb. 

Example. Find the area of the rectangle whose base is 2\ 
ft. and whose altitude is 15 in. 

Solution. 1. The formula is A = hb. 

2. h = 15 in.; b = ft., or 30 in.; A = ? 

3. .*. A = 15 X 30 = 450 sq. in. 

Note 1. — In Step 3, the symbol .*. means “therefore.” 

Note 2. — To “explain” Step 3, say, “I substituted 15 for h and 30 
for b in the formula.” Observe that 15 takes the place of h and 30 of b. 

EXERCISE 2 

1. What is the area of the rectangle whose base is 13 ft. 

and whose altitude is 8 ft. ? ( Use the form illustrated above.) 

2. Find the area of the rectangle whose base is 3 yd. and 
whose altitude is 4 ft. 

3. Find the area of the rectangle whose base is 10 ft. 6 in. 
and whose altitude is 8 ft. 

4. Find the area of the rectangle whose base is 2 rd. and 
whose altitude is 125 ft. 



4 


ALGEBRA 


x _w 

Zi l h 11 / 


7 


II. The formula for the area of a parallelogram. 

The figure XYZW is a parallelogram. 

(1) How does XYZW compare in area with 
FGHK ? 

(2) If the altitude of the parallelogram is 5 

in. and the base is 12 in., what is the area of h 
the rectangle FGHK ? What, then, must be b 

the area of the parallelogram X YZW ? 

(3) If the altitude of the parallelogram were 10 in. and the 
base 16 in., what would be its area? 

(4) Evidently the area of a parallelogram is the product of its 
base and altitude. 

If the letters h, b, and A have the meanings given on page 
3, then the formula for the area of a parallelogram is 

A = hb. 


III. The formula for the area of a triangle. 

Notice that triangle XYZ incloses one 
half as much surface as parallelogram 
XYZW. 

Since the area of the parallelogram is 
the product of its base and altitude, then 
the area of a triangle is one half the prod¬ 
uct of its base and altitude. 

Therefore the formula for the area of a triangle is 

A = \hb. 

Example. What is the area of a triangle whose base is 2-J. 
yd. and whose altitude is 2 ft. ? 

Solution. 1. The formula is A = \ hb. 

2. h = 2| yd. = 7 ft.; b = 2 ft.; A = ? 

3. Substituting in the formula, 

A = \ • 7 • 2, or 7 sq. ft. 

When solving problems by a formula, use the form illustrated 
above. 










LITERAL NUMBERS —FORMULAS 


5 


EXERCISE 3 


1. Find the area of a parallelogram whose base is 3 yd. and 
whose altitude is 4 ft. 

2. Find the area of a parallelogram whose base is 10 ft. 6 in. 
and whose altitude is 8-ft. 

3. Find the area of a triangle whose base is 3 ft. 4 in. and 
whose altitude is 15 in. (Express each in terms of inches or feet.) 

4. What is the area of a triangle whose base is 2J ft. and 
whose altitude is 20 in. ? 


6. The figure shown at the right is a rectangular solid. 
You know that its volume equals its length 
multiplied by its width, multiplied by its height. 

If the length, width, and height are represented 
by l, w, and h, respectively, and the volume by 
V, then write the formula for the volume. 


tt 

7 



- 

7 


6. Find the volume of the rectangular parallelopiped whose 
length is 30 ft., width 6 yd., and height 15 ft. 


7. a. The formula for finding the simple interest on a sum 
of money is / = PRT. Express this formula in words. 

Note. — In this formula, R is the actual rate per cent, not merely 
the number of per cent; i.e., R is 5% or .05, and not 5. 

b. Find I when P = $3600, R = 5%, and T = 4 yr. 

8. Find the simple interest on $2500 invested at 6% for 3 yr. 

9. Find the simple interest on $4600 invested at 8% for 2 yr. 
6 mo. 


10. a. The rule for finding the circumference of a circle is to 
multiply the radius by twice the number 3.1416, which is called 
pi. If r = the length of the radius, ir = the number 3.1416, 
and C = the circumference, the formula is C = 2 tt r. 

b. Find the circumference of the circle whose radius is 5 in. 


Note. — Exercise 155, page 278, can be done now. 








6 


ALGEBRA 


5. Related numbers. Every formula expresses a relation 
between two or more numbers. 

Thus, A = Kb gives the relation between A, h, and b: it tells how to 
find A when h and b are known. 

When b changes, then A changes, even if h remains fixed in value. 
Thus, if h is 5, then A = 5 b. Now, as b changes, A changes. 

This fact can be made clear by a table of their values: 


When b = 

0 

2 

4 

6 

8 

10 

then A = 

0 

10 

20 

30 

40 

50 


The relation between A and b can be shown also by a graph. 
Before doing this, we shall recall certain kinds of graphs which 
you may already know. 


6. Bar graph, a. At the right 
is a bar graph showing the amount 
earned by a certain company in 
each of certain years. 

Bar graphs are useful as means 
of presenting statistics. 

Questions. 1. How much money 
did the company earn in 1919 ? In 
1920? In 1924? 

2. Have their earnings increased 
from year to year ? 

3. In what two years did their 
earnings increase most? 

4. Make a table like the follow¬ 
ing and fill it: 


90 

80 

70 

B 60 
J3 

o 

Q 50 


s 40 
§ 

30 


20 


In the year 


10 

Their earnings were 


1919 

? 

0 

1920, etc. 

? 



1 























































































LITERAL NUMBERS — FORMULAS 


7 


b. The bars of a bar graph are often made horizontally. 

The bar graph below shows the number of customers served 
by a certain company furnishing electricity, gas, water, and ice 
in certain years. 


Number of Customers — Hundred Thousand 
123456789 10 



Questions. 1. How many customers, approximately, did 
the company serve in each year ? 

2. Approximately, how did the increases compare in the 

years from 1918 to 1922? crops 

3. When was the increase greatest ? 


c. At the right is a bar graph pic¬ 
turing the relative amounts of cer¬ 
tain agricultural products produced 
in the years 1925 and 1926. 

Questions. 1. In which year was 
the production the larger for each 
crop? 

2. For what crops was the pro¬ 
duction just about as great in 1926 
as in 1925 ? 

3. For which crop was the in¬ 
crease in production in 1926 the 
largest ? 


16,104 


17,918 


666,485 


839,818 


1,511,5 


1,282,414 g 


1,374,400 £= 


1,304,494 


325,902 


Cotton 


Wheat 


Oata 


Tobacco 


Potatoes 


360,727 











































































































8 


ALGEBRA 


EXERCISE 4 


Represent by bar graphs the following statistics. 

1. The wholesale price of eggs on the first Tuesday of each 
month of a certain year in Chicago: 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

Price per doz. 

CO 

27 i 

19(4 

18(4 

"O- 

00 

1—1 

17(4 


17?! 

22 1 

24f* 

27?! 

O 

CO 


2. The percentage of pupils studying algebra in one school, 
receiving each of certain grades: 


Grade 

96-100 

88-94 

81-87 

76-80 

In¬ 

complete 

Failed 

Per cent 

RECEIVING IT 

4 

18 

27 

26 

11 

13 


3. Tabular presentation of average retail price of butter: 


During the 
year 

1896 

1900 

1906 

1910 

1915 

1920 

Cost of 1 pound 

25(4 

26 i 

29(4 

36(4 

36(4 

67(4 


4. The population of the United States: 


Date 

1860 

1870 

1880 

1890 

1900 

1910 

1920 

Number of 
millions 

30 

39 

50 

63 

76 

92 

106 


5. The average temperature each month at Chicago: 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

Average 

temperature 

24° 

25° 

CO 

46° 

56° 

66° 

72° 

71° 

65° 

53° 

39° 

29° 









































































































LITERAL NUMBERS — FORMULAS 


9 


6. The average retail price per pound of certain articles of 
food in 1922 was : round steak, 34^; bacon, 40^; butter, 45^; 
cheese, 31^; flour, 9^; cornmeal, 4^; rice, 10^; potato, 4^; 
sugar, 7fL 

7. The average hourly velocity of the wind in miles per 
hour at certain points in the United States was: 

Boise, Idaho . . 4 mi. per hr. Portland, Me. . 


Atlanta, Ga. 
Buffalo, N.Y. 
Duluth, Minn. 
Omaha, Neb. 


9 mi. per hr. 
11 mi. per hr. 

7 mi. per hr. 

8 mi. per hr. 


Vicksburg, Miss. 
Philadelphia, Pa. 
Chicago, Ill. 
Boston, Mass. 


5 mi. per hr. 
16 mi. per hr. 

10 mi. per hr. 
9 mi. per hr. 

11 mi. per hr. 


8 . The average number of inches of precipitation (rain and 
snow) at certain cities is : 


San Diego, Cal. . . 

. . 10 in. 

Mobile, Ala. . . . 

62 in. 

Denver, Colo. . . 

. . 14 in. 

Flagstaff, Ariz. . . . 

23 in. 

Springfield, Ill. . . 

. . 37 in. 

Salt Lake City, Utah 

. 16 in. 

Dubuque, la. . . . 


Wilmington, N.C. 

51 in. 

Baltimore, Md. . . 

. . 43 in. 

Portland, Me. . . . 

. 42.5 in. 

• 9 . The number of days required for the germination of 

certain seeds is: 

Bean. 

7 da. 

Lettuce. 

7 da. 

Beet. 

. . 8.5 da. 

Onion. 

. 8.5 da. 

Carrot. 


Pea. 


Corn . 

. . 6.5 da. 

Radish . 

. 4.5 da. 


10. The lengths of certain rivers: Mississippi, 3000 mi.; 
St. Lawrence, 2200 mi.; Yukon, 2100 mi.; Arkansas, 2000 mi.; 
San Francisco, 1400 mi.; Columbia, 1200 mi. 

11 . The percentages of illiterates in 1920 in the various 


sections of the United States were: 


New England. 

4.9 

East South Central . . 

. 12.7 

Middle Atlantic .... 

4.9 

West South Central ; . 

. 10.0 

East North Central . . . 

2.9 

Mountain. 

. 5.2 

West North Central . . . 

2.0 

Pacific. 

. 2,7 

South Atlantic. 

11.0 





















10 


ALGEBRA 


7 . Representing statistics by broken line graphs. 

Example. The average number of pounds of sugar which 
could be bought for $1 in certain years is given below. 


Year 

1913 

1914 

1915 

1916 

1917 

1918 

1919 

1920 

1921 

1922 

No. of Lb. 

18 

17 

15 

m 

11 

10 

9 

5 

m 

14 


Explanation of the graph which appears below. 

On the heavy horizontal line, called the horizontal axis, 
points representing the years are spaced at equal distances, 
and are marked by the number of years. The distance between 
two points is determined by the size of the piece of graph paper. 


Pounds 



On the heavy vertical line, call the vertical axis, points repre¬ 
senting 0, 5, 10, 15, and 20 pounds are placed at equal distances, 
and marked. One small space was taken to represent 1 pound. 

Above each point representing a year, a dot is placed opposite 
the number of pounds of sugar which $1 would buy in that year. 
Thus R, above 1915, is opposite 15. This means that $1 would 
buy 15 pounds in 1915. In this manner a dot was placed above 
each year point. 

Next, each dot was connected with the next one by a straight 
line. The complete line is called a broken line. 

























































































LITERAL NUMBERS — FORMULAS 


11 


EXERCISE 5 


Represent by broken-line graphs the following statistics : 


1. Year 

1850 

I860 

1870 

1880 

1890 

1900 

1910 

1920 

Population of United 
States per sq. mi. 

8 

11 

13 

17 

21 

26 

31 

36 


2. Hour oe Day 

6 

7 

8 

9 

10 

11 

12 

l 

2 

3 

4 

5 

Temperature 

12° 

14° 

18° 

© 

o 

o 

CO 

fcO 

0 

33° 

35° 

32° 

o 

o 

CO 

26° 

25° 


3. Month of Year 

Oct. 

Nov. 

Dec. 

Jan. 

Feb. 

Mar. 

Apr. 

May 

Average per cent of 
year’s supply of coal 

BURNED 

7 

11 

17 

18 

17 

14 

10 

6 


4. Year 

1913 

1914 

1915 

1916 

1917 

1918 

1919 

1920 

1921 

1922 

NO. OF LB. OF ROUND 
STEAK $1 WOULD BUY 

4.5 

4.2 

4.3 

4.1 

3.4 

2.7 

2.6 

2.5 

2.9 

3.0 


5. Year 

1913 

1914 

1915 

1916 

1917 

1918 

1919 

1920 

1921 

1922 

No. OF LB. OF SIRLOIN 
STEAK $1 WOULD BUY 

3.9 

3.9 

3.9 

3.7 

3.2 

2.6 

2.4 

2.3 

2.6 

2.5 


6. Below are given the numbers of deaths caused by auto¬ 
mobiles in two cities of New York, in recent years. Represent 
the numbers for one city on one broken-line graph, and those 
for the other city by a second graph on the same set of axes. 


Year 

1915 

1916 

1917 

1918 

1919 

1920 

City A 

42 

56 

81 

90 

68 

104 

City B 

18 

29 

21 

25 

32 

34 







































































































12 


ALGEBRA 


8. Representing statistics by smooth-line graphs. 

Example. The temperature readings at certain hours : 


7 A.M. + 2° 

8 A.M. + 4 ° 

9 A.M. + 7° 


10 A.M. + 10° 

11 A.M. + 14° 

12 m. + 16° 


1 P.M. + 17° 

2 P.M. + 16° 

3 P.M. + 14° 


4 P.M. + 12° 

5 p.m. + 10° 

6 p.m. 8 ° 


The plus sign means that the temperature was above zero. 
Explanation of the graph below. 

On the vertical axis, observe the points marked + 5°, + 10°, 
and + 15°. 

On the horizontal axis, observe the point marked 7, 8, 9, etc., 
representing the hours of the day. 

Points were placed above the hour points, and opposite the 
temperature points. Thus, above 8 a.m. a dot is found opposite 
+ 4, because the temperature was + 4° at 8 a.m. 

These points were connected by a smooth curved line. 



This graph enables us to arrive at new information. Of 

course the temperature did not jump from 10° at 10 a.m. to 
14° at 11 a.m. It probably rose gradually. From the graph 
we can find the probable approximate temperature at, say, 
10 : 40 a.m. The point S indicates the time 10 : 40. On the 
graph, above S, is point T. Point T is opposite 12.5°. We 
conclude that the temperature at 10 : 40 was about + 12.5°. 











































































































LITERAL NUMBERS —FORMULAS 


13 


EXERCISE 6 


Represent the following sets of statistics by curved-line graphs. 


1. Hour 

6 A.M. 

7 A.M. 

8 A.M. 

9 A.M. 

10 A.M. 

11 A.M. 

Temperature 

+ 12° 

+ 13° 

+ 14° 

+ 15° 

+ 17° 

+ 18° 


Hour 

12 M. 

1 P.M. 

2 P.M. 

3 P.M. 

■4 P.M. 

5 P.M. 

Temperature 

+ 18° 

+ 16° 

0 

o 

yH 

+ 

+ 5° 

+ 3° 

0° 


Find the approximate temperature at 1 : 30 and 4:30 p.m. 

2. The amount of electricity used by one family in one year : 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

Kilowatt-hours 

27 

27 

24 

19 

15 

9 

10 

8 

18 

42 

49 

45 


3. The amount of gas used by one family in a year: 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

No. of 100 

CU. FT. USED 

36 

32 

29 

31 

29 

1 

29 

23 

22 

31 

32 

28 

22 


4. The amounts collected by each of two teams in one week 
of a “ drive ” for charitable purposes. Represent the collec¬ 
tions by two graphs on the same set of axes, one for each team. 


Day 

Mon. 

Tues. 

Wed. 

Thurs. 

Fri. 

Sat. 

Team A 

Team B 

$108 

132 

$ 92 

140 

$ 76 

105 

$ 87 
150 

$108 

175 

$125 

190 


















































































14 


ALGEBRA 


9. Representing a formula by a line graph. 

The area of the rectangle having base 6 and altitude 5 is 
given by the formula H = 5 6. This relation is represented in 
the graph below. 


How the graph is made and used 


The graph 


1. First make a table as follows 


When 6 = 

0 

2 

4 

6 

8 

10 

then A = 

0 

10 

20 

30 

40 

50 


2. OA and OB are ruled heavily 
on coordinate paper. 

OB will be called the b line. 

OA will be called the A line. 

On OB, 2 spaces represent 1 unit. 

On OA, 1 space represents 2 units. 

3. When b = 2, A = 10. Above 
2 on the b line, point E is placed 
opposite 10 on the A line. 

When b = 4, A = 20. Above 4 on the b line, point F is 
placed opposite 20 on the A line. 

Similarly, points 0, G, H, and I are located. 

4. The line joining 0, E, F, G, etc., is the graph of the formula 
A = 5 6. 



123456789 10 


5. To find how much A is when 6 = 3, start at b = 3; move 
up to the graph, arriving at point R ; move to the left, arriving 
at the A line, at the point which represents 15. Therefore 
A = 15, when 6=3. 

6. To find how much 6 is when A = 35, start at 35 on the 
A line. Move to the right, arriving at S on the graph, now 
move down to the 6 line, arriving at the point which represents 
6 = 7. Therefore 6 = 7 when A = 35. 



































































































LITERAL NUMBERS — FORMULAS 


15 


EXERCISE 7 

1. a. Draw a graph of the formula ^4 = 46; when 6 = 0, 
2, 4, 6, 8, 10, and 12. 

6. Find A when 6 = 9. 

c. Find 6 when A = 22. 

d. When 6 increases, what happens to A ? 

2. The area of a triangle with base 6 and altitude 7 is given 
by the formula A = f X 7 X 6, or ^4 = -J 6. 

а. Make a table of values of ^1 when b = 0,2, 4, 6, 8, and 10. 

б. Draw the graph. 

c. From your graph, determine A when 6 = 5. 

d. When A increases, what happens to 6 ? 

3. The formula expressing the cost, C, of n pounds of any 
article at 32^ per pound is C = 32 nj. 

а. Make a table of values of C when n = 0, 1, 2, 3, 4, and 5. 

б. Draw the graph using the horizontal axis for the value of n. 
(Let 1 space represent f lb.; vertically, let 1 space represent 4^.) 

c. From your graph, find the cost of 2\ lb.; of 3f lb.; of If lb. 

d. How does C change when n increases ? 

4. a. Write the formula expressing the total wages, W, of a 
workman who works n hours at 75per hour. 

6. Represent this relation graphically, for n = 0, 8, 1G, 24, 
32, 40, 44. 

c. Find W when n = 21; when n = 36; when n = 18. 

d. When n is doubled, what happens to W ? 

6. a. The formula expressing the interest, I, on $100 at 
6%, for n years is I = 6 n. 

6. Make a table of values of I when n — 1, 2, 3, . . ... 10. 
Draw the graph, using the horizontal axis for the values of n. 

c. From your graph, determine the interest on $100 at 6% 
when n = 3f yr.; when n = 5f yr.; when n = yr. 

d. How does I change when n increases ? 

e. How does I change when n is doubled ? 


16 


ALGEBRA 


10. Multiplication and division of a product by an arithmetical 
number. 

In arithmetic, 2 X 3 X 4 =(2 X 3)X 4 = 6 X 4 = 24. 
Similarly, in algebra, 5X4 m = (5 X 4) X m, or 20 m. 

Note. — The symbols ( ) are called parentheses. The numbers inside 
are to be combined before using them in any other arithmetical manner. 

Since 2 • 3 a = 6 a, then 6 a -f- 2 =3 a. 

Similarly, 20 a; 5 = 4 a;; 18 y ^ 9 = 2 y. 


EXERCISE 8 

Give the following products orally, or write only the results 


7 - 8® 

3 • 11 b 
9-12 m 

8 • 15* 


12 -8 w 
11-7 t 
6 • 13 v 


9. 

10 . 

11 . 

12 . 


4 - fa; 

5- *y 

6- f* 
12-4i 


13. 20 • f m 

14. 24 • f p 

15. 36 • % r 


8. 9 • 15 3 

Give orally or write only the quotients 

17. 25 t "v" 5 21. 36 x -r- 36 

18. 32 r -7- 8 22. 29 y -f- 29 

19. 60 w -f- 12 23. 39 t ^ 39 

20. 24 x -s- 4 24. 2.5 2 -r- 2.5 

29. What part of 32 a; is : a. 8#? h. 

30. What part of 28 t is : a. 2 * ? b. 

31. a. By what must you divide 6 x to get x ? 

b. By what must you divide 18 r to get 2 r ? 

c. By what must you divide 25 z to get z ? 

d. By what must you divide 3.5 x to get x? 

Give the following quotients : 


84 p + 12 
16 m -j- 16 
.5 w -r- .5 
i « x 


25. 

26. 

27. 

28. ^z _ 

16 z? c. x? 
7*? c. *? 


32. 


34. 


y + 


36. im + i 

37. Y < - ¥ 

38. 2.3 z ^ 2.3 

39. 4.5 r ^ 4.5 


40. 

41. 

42. 

43. 


2.37 y ^ 2.37 
14.8 z ^ 14.8 
ax -T- a 
mx -7- m 


LITERAL NUMBERS — FORMULAS 


17 


11. An equation expresses an equality of numbers, 
formula is one kind of equation. 

Thus, A = 3 b is an equation. 

The numbers on the right side of the equality sign form 
right side of the equation ; the ones on the left side of 
equality sign form the left side of the equation. 

12. Solving an equation. 

Consider the equation 3 x + 11 = 26. 

When x = 5, 3 * +- H is 3 • 5 + 11 or 15 + 11. 

This does indeed equal 26. 5 is said to satisfy the equation. 

5 is called a root of the equation. 

But when x is any other number, 3 x +- 11 does not equal 26. 

Thus, if x = 8, 3 • 8 + 11 = 24 + 11, or 35. 8 does not satisfy 
the equation, because 3-8 + 11 does not equal 26. 

An equation like 3 x + 11 = 26 is satisfied by only one value 
of x. The process of finding that value is called solving the equation. 

Thus, 6 satisfies the equation 3 x + 8 = 26 for 18 + 8 = 26. 

3 does not satisfy the equation 5 2 + 11 = 31 for 15 + 11 does not 
equal 31. 

13. Rule I used in solving equations. Suppose the weight 
and the sugar just balance on the scales 
pictured at the right. 

Then one half the sugar will also balance 
one half the weight. One half the sugar will not balance one 
third of the weight. 

Similarly, in an equation like 10 x = 80, if both sides are 
divided by 2, the results must be equal : 
that is, 5 x = 40. 

But if one side is divided by 2, and the other side is divided 
by some other number, the results will not be equal. 

Rule I. Both members of an equation may be divided by 
the same number without destroying the equality. 



A 

the 

the 










18 


ALGEBRA 


14. Solving equations by use of Rule I. 

Example 1. Solve the equation 17 y = 391. 

Solution. 1. y can be obtained from 17 y by dividing 
17 y by 17. Then, in order to keep the two sides equal, 
both sides must be divided by 17. 

2. Then, y - 23. 


23 

17)391 

34 

51 

51 


In order to abbreviate the written explanations of solutions 
of equations, the symbol D will be used as follows : 

Di 7 will mean “ divide both sides of the preceding equation 

by 17.” 

Example 2. Solve the equation f x = 18. 


Solution. 1. f a: = 18 

2. D| x — 18 -r | 

3. * = 24 


f X £ = X 

3 6 4 

18 + | =J^X| 

= 24 


Note. — In Step 2, what does Dj mean? 


EXERCISE 9 


Solve the following equations, using the form of solution 
illustrated in Example 2 : 


1. 

7 x = 

21 

8. 

15 = 

: 3 w 

15. 

15 x = 60 

2. 

4 y = 

28 

9. 

26 = 

■■ 2 r 

16. 

21 = 7 M 

3. 

.5 z = 

20 

10. 

44 = 

■ 4 s 

17. 

20 y = 40 

4. 

6 m = 

: 24 

11. 

12 x 

= 36 

18. 

22 = 11 

5. 

St = 

48 

12. 

45 = 

: 9 b 

19. 

4x = 18 

6. 

9 A = 

= 63 

13. 

14 y 

= 42 

20. 

II 

5ss 

CO 

7. 

10 c- 

= 90 

14. 

65 = 

: 13 2 

21. 

5t = 14 

22. 

^ x = 

2 

25. 

i9 = 

= 11 

28. 

iA = 10 

23. 

iv = 

5 

26. 

ix -- 

= 1 

29. 

4 = in 

24. 


7 

27. 

3 = 


30. 

7 ~ i z 

31. 

If x ~ 

40 

34. 

9 = 

I 'W 

37. 

-y- C = 99 

32. 

A f = 
3 1 

16 

35. 

12 = 

■- %x 

38. 

# a = 120 

33. 

iw = 

= 14 

36. 

10 = 

-iC 

39. 

f m = 189 




LITERAL NUMBERS — FORMULAS 


19 


15. Solving problems by means of equations. 

Use the following form of solution for the problems in Exercise 10, 
even though the arithmetical solution appears easier at this time. 

EXERCISE 10, a 

1. If a certain number be multiplied by 17, the product is 

144.5. What is the number? 

Solution. 1. Let n = the number. 

2. Then 17 n = 17 times the number. 

3. Then 17 n = 144.5, according to the statement of the 

problem. 

Finish this problem by dividing both sides by 17. 

2. If a certain number be multiplied by 13, the product is 
247. What is the number ? 

3. If a certain number be multiplied by 8.5, the result is 221. 
What is the number? 

4. How many articles at $1.75 each can be bought for $40.25 ? 
Solution. 1. Let n = the number of articles. 

2. n • $1.75 or $1.75 n = the cost of the articles. 

3. 1.75 n = 40.25. 

Complete this problem. 

6. How many articles at $2.15 each can be bought for $38.70 ? 

6. What number, multiplied by 2.7, gives 9.45 as product? 

7. How many miles per hour did an auto travel which went 
172 miles in 8 hours ? 

8. At what price must an article be purchased if 16 of them 
cost $13.60? 

9. If a certain number be multiplied by 12.5, the product 
is 217.25. What is the number ? 

10. How much must each of 25 persons contribute to make 
up a fund of $156.25, if all contribute equally? 

11 . The product of two numbers is 499.5. One of them is 

18.5. Whatis the other? 


20 


ALGEBRA 


EXERCISE 10, b 

1. Fifteen ninths of a certain number is 75. What is the 
number ? 

Solution. 1. Let n = the number. 

2. is.n = 75. 

Complete the solution. 

2. Five ninths of a certain number is 55. What is the 
number ? 

3. Seven thirds of a certain number is 112. What is the 
number ? 

4. Eight fifths of the cost of a certain lot was $3000. What 
was the cost of the lot ? 

6. Three eighths of the area of Lake Michigan is 8625 
sq. mi. What is the area of Lake Michigan ? 

6. When asked her age, a girl replied, seven fifths of it is 21. 
What was her age ? 

7. Four sevenths of the members of a certain class were 
boys. If there were 28 boys, what was the total membership 
of the class ? 

8. The selling price of a certain article was five thirds 
of its cost. What must have been its cost if it sold for $3.50? 

9. Five eighths of the cost of the Suez Canal was $62,500,000. 
What was the cost of the canal ? 

10. Seven eighths of the population of New York in 1920 was 
9,100,000. Find the population in 1920. 

11. Nine tenths of the cost of an automobile tire was $24.75. 
What was the cost of the tire ? 

12. 1.25 times the cost of a certain article was $2.00. What 
was the cost of the article ? 

13. 130% of the cost of a certain article was $4.55, What 
was the cost of the article ? 


LITERAL NUMBERS — FORMULAS 


21 


16. An important law of multiplication. 

You know that 2 X 3 X 4 = 6 X 4, or 24; 

that 3 X 2 X 4 = 6 X 4, or 24; 

that 2 X 4 X 3 = 8 X 3, or 24. 

That is, 2X3X4 = 3X2X4 = 2X4X3. 

Rule. — The factors of a product may be rearranged in any 
order before finding the product. 

Example 1. 6 • a • 5 = 6 • 5 • a, or 30 a. 

2 

Example 2. 8 • b • - = $ • - • 6, or 6 b. 

4 i 


EXERCISE 11 

Find mentally the following indicated products : 

1. 7 • m • 8 4. 12 • y • f 7. 2 • r • 5 • 3 

2. 6 • y • 9 5. 16 • x • -J 8. 6 • s • 4 • 5 

3. 7 • z • 3 6. 18 • 2 • | 9. 15 • t • % • 8 

17. Finding any number in a simple formula. 


Example 1. From the formula V = Iwh, find w if 
V = 1456, l = 7, and h = 16. 
Solution. 1. The formula is V = Iwh. 

V = 1456; l = 7; h = 16; = ? 

2. Substituting, 1456 = 7 • w • 16. 

3. .*. 1456 = 112u>, 

or 11210 = 1456. 

4. Dn2 w = 13. 


Example 2. 

Solution. 1. 
or 

2. D ¥ 

3. 

4. 


From the formula V = -g- find 6 when 
F = 117 and A = 13. 

Substituting, 117 = £ • 13 • 6, 

= 117. 


6 = 117 -T- 


6 = 27. 


22 


ALGEBRA 


EXERCISE 12 

By the formula A = hb : 

1. Find b, when A = 90 and h = 10. 

2. Find h, when b = 15 and A = 120. 

3. Find A, when h = 6 and b = 14. 

4. Find A, when h = 12 and b = 28. 

5. How does A change if both h and b increase ? 

6. How does A change if h is doubled and b is trebled ? 

By the formula V = Iwh : 

7. Find h, when V = 480, l = 12, and w — 8. 

8. Find w, when V = 1560, l = 20, and h = 13. 

9. Find V, when l = 15, w = 10, and h = 5. 

10. Find V, when l = 15, w = 10, and h = 15. 

11. Find l, when w = 10, h — 15, and V = 1800. 

12. How does V change if l and w have fixed values and h 
increases ? 

13. What is the effect on V if w and h have fixed values and 
l is doubled ? 

By the formula A = hb: 

14. Find b, when A = 189 and h = 14. 

15. Find h, when b = 22 and A = 154. 

16. Find h, when b = 22 and A = 88. 

17. How does h change if b has a fixed value and A decreases ? 
By the formula V = ^ hb: 

18. Find b, when V = 57 and h = 19. 

19. Find h, when V = 232 and b = 29. 

20. Find h, when V = 232 and b = 24. 

21. How does h change if V has a fixed value and b decreases ? 
By the formula I = PR T : 

22. Find T, when I = $273, P = $650, and R = 7%. 

23. Find P, when 7 = $555, R = 6%, and T = 2 yr. 

24. Find P, when 7 = $375, T = 1-J yr., and R = 4%. 

Note. — Exercise 156, p. 279, can be done now. 


LITERAL NUMBERS — FORMULAS 


23 


ALGEBRAIC EXPRESSIONS 

18. An algebraic expression is a number expressed by literal 
and arithmetical numbers, connected by mathematical signs 
which tell what to do with the numbers. 

Thus, 2 x — 3 yz + z is an expression. 

A monomial or term consists of numbers connected only by 
signs of multiplication or division; as, 2 x 2 , or 3 ab. 

The parts of an expression separated by plus or minus signs are 
terms. Thus, 2 x, 3 yz , and 2 are the terms of 2 x — 3 zy + z. 

19. In a term like 2 a;, 2 is called the numerical coefficient of 
x. Similarly 3 is the numerical coefficient of zy. z is the same 
as 1 z, therefore the numerical coefficient of z is 1. 

20. The arithmetical or numerical value of an expression is 
found by substituting arithmetical values for the literal num¬ 
bers, and doing what the signs direct. 

21. Order of fundamental operations. Addition, subtraction, 
multiplication, and division are called the four fundamental opera¬ 
tions. Often several of them must be performed when finding 
the value of an expression, — as in the examples above. 

Rule. —To find the value of an expression: 

1. First, do all the multiplications. 

2. Second, do all the divisions, taking them in order from 
left to right. 

3. Then do the additions and subtractions, taking them in 
any order. 

Thus, if x = 15, y — 2, and z = 3 in 2 x — 3 yz, 
then 2 z - 3 yz = 2 X 15 —3X2X3 
= 30 - 18, or 12. 

If, instead, we had first subtracted 3 from 15, getting 12; had mul¬ 
tiplied this by 2, getting 24; had multiplied this by 2 and by 3, the final 
result would have been 144. The correct result is 12. 


24 


ALGEBRA 


EXERCISE 13 

Find the numerical value of the following expressions when 
a = 2, b = 4, c = 0, d = 3, x = 2, y = 5: 


1 

a + b 

6. 

y — x + 6 

11. 

3 a -f- b — 5 

2. 

c — X 

7. 

ax + by 

12. 

2c- 

■ 3 a + d 

3. 

6 d + y 

8. 

cd — bx 

13. 

ab + 

,cd — xy 

4. 

3 c + 2 d 

9. 

3 ab + 2 cy 

14. 

bd- 

ay + cx 

5. 

ab — d 

10. 

abc — dxy 

15. 

by - 

ax — cd 


Write an algebraic expression to represent: 

16. The sum of a and 5; of a and b. 

17. The product of c and 4; of c and d. 

18. The quotient of 15 divided by a;; of y divided by x. 

19. The product of a, x, and y. 

20. The product of c and d, divided by a. 

21. 8 more than x ; 10 more than 3 y. 

22. 7 times the product of c and d. 

23. One half the product of a and b. 

24. One third the product of H and B. 

25. The sum of A and P. 

26. A increased by B ; A decreased by B ; A diminished by B. 

27. The sum of twice x and three times x. 

28. The sum of n, 3 times n, and 5 times n. 

29. The sum of 13 and ■§■ ra. 

30. One number is represented by n. 

a. Represent the number which is 5 times as large. 

b. Represent the sum of these two numbers. 

31. One box weighs y pounds. A second weighs -J as much 
as the first, and a third weighs § as much as the first. 

a. Represent the weight of the second and the third boxes. 

b. Represent the sum of the weights of the three boxes. 

32. Charles’ age is x years. Edward is twice as old as 
Charles, and John is f as old as Charles. Represent the age 
of Edward; also of John. 


LITERAL NUMBERS — FORMULAS 


25 


EXERCISE 14 

1. The sum of two numbers is 18. One of them is z. 
What is the other ? 

2. The sum of two numbers is N. One of them is 15. 
What is the other ? 

3. The sum of two numbers is S. One of them is n. 
What is the other ? 

4. What was the selling price of an article bought for p 
dollars and sold at a gain of $5 ? Bought for C dollars and 
sold at a gain of G dollars ? 

5. What is the weight of 25 bags of grain, each weighing 
x pounds ? Of n bags, each weighing t pounds ? 

6. What is the cost of 5 gallons of gasoline at 23^ per 
gallon ? Of m gallons at y cents per gallon ? 

7. a. What is the total weight of n chickens, weighing, 
on the average, r pounds each ? 

b. What is their total value, at 20j£ per pound ? At c cents 
per pound ? 

8. How much change should be received from a $5 bill, 
offered in payment for 4 yards of cloth at 65^ per yard ? For 
m yards at 50^ per yard ? For k yards at r cents per yard ? 

9. What is the total cost of n pounds of coffee at x cents 
per pound and m pounds at y cents per pound ? 

10. What is the cost of 18 eggs at the rate of 75^ per dozen ? 
Of n eggs at the rate of y cents per dozen ? 

11. How many quarts are there in 8 gallons ? In t gallons ? 
In p gallons and 2 quarts ? In r gallons and s quarts ? 

12. How many cents are there in z dollars ? In x dollars 
and 4 dimes ? In y dollars and n dimes ? 

13. How much remains when a piece a feet long is cut from 
a board 10 feet long ? When a piece d feet long is cut from a 
board m feet long ? 


26 


ALGEBRA 


22. Adding and subtracting numbers having a common factor. 
A factor of a number is an exact divisor of it. 

A common factor of two or more numbers is a factor of each 
of them. 

Thus, 3 is a common factor of 6 and 9. 

x is a common factor of 4 x and 11 x. 

Like terms are numbers having a common literal factor. 

Thus, 4 x and 11 x are like terms. 

The result obtained by adding two or more numbers is called 
their sum. The numbers themselves are called addends. 


Just as 3 times 7 + 2 times 7 = 5 times 7, 

so 3 times x + 2 times x = 5 times x. 

Similarly 5a + 3a = 8a; and 6 ra + 4 m = 10 m; 

also 5a — 3 a = 2 a; and 6 m — 5 m = m. 


Adding or subtracting like terms is also called combining like 
terms. 


EXERCISE 15 


Combine the terms in the following examples: 


1. 3 

x + 4 x = 

6. 

12 a: 

— 4 x = 


11. 

14 y - 6 y = 

2. 5 

c + 2 c = 

7. 

15 y 

- y = 


12. 

7t + 13 t = 

3. 7 

m + m = 

8. 

11 1 - 

-5 t = 


13. 

18i/- 12 7/ = 

4. r 

+ 5 r = 

9. 

9 m - 

— m = 


14. 

9 w + 12 w = 

6. 6 

p~h 9 p = 

10. 

17 a 

— 9 a — 


15. 

15 2 — 6 2 = 

16. 

10x + 5 x + c 

1 x = 


22. 

4s 

: + * 

— 52 = 

17. 

2 y + 3 y + y 

= 


23. 

11 

y - 

6 7/ + 9y = 

18. 

5 to + 13 m — 

m - 


24. 

15 

t — ; 

t- 3t = 

19. 

16 a — a+9a = 


25. 

g+5g 

-3 g = 

20. 

7 2 + 12 2 + 2 

= 


26. 

12 

a — 

9 a — a = 

21. 

2 x + 6 x — 3 

x = 


27. 

2+10 

2 — 2 = 

28. f 

t + ±t = 

31. 

5 o _ 

6 S 

’ i 8 ~ 


34. 

5 c + .2 c = 

29. i 

r + ir = 

32. 

iy- 

" To v — 


35. 

x + .25 a; = 

30. i 

t + i t = 

33. 


- f m = 


36. 

8.5 7 / - 2.3 7 / = 


LITERAL NUMBERS —FORMULAS 


27 


23. Combining like terms used in solving equations. 

Example. Solve the equation 4 m — m + 11m = 112. 
Solution. 1. 4m-m + llm = 112. 

2. Combining terms 14 m = 112. 

3. I) 1 4 m = 8. 

Check. Substituting 8 for m in the original equation, does 
32 - 8 + 88 = 112? Does 24 + 88 = 112? Yes. 


EXERCISE 16, a 


Solve and check the following equations: 


1. 5 r + 3 r = 96 

2. 7 t + 4 t — 88 

3. 9 x + 3 x = 72 

4. 12 m + 3 m = 105 

5. 6 y + 7 y = 65 

6. 12 x — 4 x = 88 

7. 13 y — 7 y = 54 

8. 17 m — 5 m = 96 

9. 18 s — 7 s = 99 

10. 16 c — 9 c = 84 

11. 11 A + 5 A = 96 

12. 15 G - 8 G = 98 

13. 12 x + 13 x = 175 

14. 23 ?/ — 9 2 / = 112 

15. 17 t + 14 t = 155 

EXERCISE 


16. 3ic + 4ic + 5x = 120 

17. 7 y — y + 2 y = 136 

18. 9 r + 2 r — 5r = 138 

19. £ + 8/ +7£ = 208 

20. 15 z — 3 3 — z = 198 

21. 2 x + 6 x — 3 x = 115 

22. 11 y — 6 y + 9 y = 22+ 

23. 5c+9c-6c =+4t38 

24. 11 w — w + 12 w = 264 

25. s + 15 s — 4 fs = 75 

26. 8 a — a — y a = 7 

27. 6 P + 5 P - P = 42 

28. 15 t + t - 3 t = 273 

29. 22 ^ - 5 + 6 5 = 90 

30. 15 x — 3 x + 5 x = 68 

16, b 


Speed and accuracy test. Time: 5 minutes. 
Solve and check the following equations: 


1. 8 x + 3 x — 77 

2. 13 z - 4 z = 90 

3. 5 y -f 11 y = 80 

4. 20t — St= 108 

5. 3 w + 15 w = 198 


6. 2x-\-Sx-{-5x = 110 

7. 6 y + 5 y - 3 y = 120 

8. 4z — 3z + 6z = 105 

9. 12 t — 5 t + 3 t = 80 
10. w 11 w — 4w = 104 


28 


ALGEBRA 


24. Addition and subtraction of literal numbers used in 
solving problems by means of equations. 

Example. The sum of two numbers is 96. The larger 
number is 11 times the smaller. What are the numbers ? 
Solution. 1. Let s = the smaller number. 

2. Then 11 s = the larger number. 

3. Then s + 11 s = the sum of these two numbers. 

4. .*. s + 11 s = 96, since the sum of the numbers is 96. 

5. Adding, 12 s = 96. 

6. D 12 s = 8, the smaller number. 

7. .‘.118 = 88, the larger number. 

Check. Is 96 the sum of 8 and 88? Yes. 

Is the larger, 88, eleven times the smaller, 8? Yes. 


EXERCISE 17 

1. The greater of two numbers is 5 times the smaller. The 
sum of the two numbers is 144. What are the numbers ? 

2. The sum of two numbers is 135. The greater is 14 times 
the smaller. What are the numbers ? 

3. The greater of two numbers is 7 times the smaller. The 
greater less the smaller is 48. What are the numbers ? 

4. The larger of two numbers is 10 times the smaller. The 
larger diminished by the smaller is 67.5. What are the numbers ? 

5. James’ age is 5 times Fred’s age. The sum of their ages 
is 54 years. How old is each ? 

6. A man owns two farms having a total of 225 acres. The 
second farm is 4 times as large as the first. How large is each 
farm ? 

7. Two boys made a profit of $3.80 from selling lemonade. 
They had agreed that the second should receive 3 times as much 
as the first. How much should each receive ? 

8. The second of three numbers is 3 times the first; the third 
is 5 times the first. The sum of the three numbers is 126. What 
are the numbers ? 


LITERAL NUMBERS — FORMULAS 


29 


9. The sum of three numbers is 480. The second is 4 times 
the first, and the third is 5 times the first. What are the 
numbers ? 

10. The sum of three numbers is 260. The second is 3 times 
the first; the third is 3 times the second. What are the 
numbers ? 

11 . Separate 69 into two parts such that the larger is 22 
times the smaller. 

12. Separate 18.6 into two parts such that the larger is 5 
times the smaller. 

13. Separate 108 into three parts such that the second is 

3 times the first, and the third is 5 times the first. 

14. Separate 156 into three parts such that the second is 

4 times the first, and the third is 7 times the first. 

15. Divide $175 amGng three children so that Charles shall 
receive twice as much as John, and Mary twice as much as 
Charles. 

16. The length of a lot is 4 times its width. The distance 
around it is 350 ft. What is its width and length ? 

17. The total weight of three boxes is 500 lb. The second 
weighs three fourths as much as the first, and the third five 
fourths as much as the first. What does each weigh ? 

18. In order to raise a fund of $1000, A agrees to give 3 
times as much as B, and C to give twice as much as A. How 
much must each give ? 

19. A man agreed to give a certain amount to a fund pro¬ 
vided a second man gave twice as much. A third man agreed 
to give twice as much as the second provided they made a total 
contribution of $350. How much did each have to contribute ? 

20. Separate 285 into four parts such that the second is 3 
times the first; the third twice the first; and the fourth 3 
times the second. 


30 


ALGEBRA 


MAKING AND USING MORE DIFFICULT FORMULAS 

25. A rule of computation expresses in words the relation of one 
number to one or more other numbers. 

A formula expresses the same relation by means of algebraic 
symbols. 

To do this more algebraic symbols and words must be learned. 

a. Parentheses, ( ), are used to inclose a part of an expression 

which must be treated as a single number. 

Thus i(9 + 7) means that 9 and 7 must be added, and that £ of the 
result must then be found. This gives 8 as the final result. 

b. Power, base, and exponent. When the same number is 
used as a factor several times, the result is called a power of the 
number; the number itself is called the base; a small integer 
written at the right of and above the base tells how many times 
the base is used as a factor and is called an exponent. 

Thus x 2 means x • x. Read it “x square” or “ x second power.” 
x 3 means x • x • x. Read it “x cube” or “x third power.” 
x 4 means x • x • x • x. Read it “x fourth” or “x fourth power.” 

The number x is the base in each of these illustrations; the numbers 
2, 3, and 4 are exponents. 


EXERCISE 18 

1. a. How much is x 2 when x = 2? a: = 3 ? x = 4? 

b. How does x 2 change when x is more than 1 and increases ? 

2 . How much is x 2 when & = ■£.? a; = -§-? x = 

3. How much is x 3 when £ = 2? x = 3? x = 4? 

4. How much is y 3 when y = 1 ? y = -J-? y = ■|? 

5. How much is (a) (l) 6 ? (6) (2) 4 ? '(c) (-J) 3 ? 

6. (a) Complete the following table in which y = x 2 . 

(b) What change takes place 

in y when x is doubled ? 

7. Which of the following 
statements are true and which 

false: (a) 2 3 = 6? (b) 0 4 = 0? (c) 2 4 = 8? (d) 3 4 = 27? 


When x = 

1 

2 

4 

8 

then y = 

? 

? 

? 

? 








LITERAL NUMBERS — FORMULAS 


31 


8. In simple interest problems, the amount equals the prin¬ 
cipal plus the interest. 

a. Express this relation as a formula, using A for the amount, 
P for the principal, and I for the interest. 

b. Using this formula, find A when P = $600 and I = $16.50. 

c. Also find A when P = $800 and I = $25. 

d. If P remains the same in value, but I increases, what 
happens to A ? 

e. If I remains the same in value and P increases, what 
happens to A? 

9. The selling price of an article equals its cost plus the gain 
made when selling it. 

a. Express this relation between the selling price, S, the cost, 
C, and the gain, G. 

b. Using your formula, find S when C = $6.50 and G = $2.25. 

c. If C has a fixed value, how does S change if G increases? 

d. If C has a fixed value, how does G change if 8 increases ? 

10. You know that the area of a circle equals pi (3.1416 or 3^-) 
times the square of the radius. 

a. Express this rule as a formula, using A for the area, 7r for 
the number pi, and r for the radius. 

b. Find, by your formula, the area, A, when r = 7. 

c. Find the area of the circle whose diameter is 16 in. 

d. Complete the following table, using 7r = 3.1. 

e. Observing the results in 
(d), if r is doubled, then A 

is (?) 

11. It is proved in geometry that the volume, V, of a sphere 
equals four thirds of pi times the cube of the radius, r. 

a. Express this relation between V, i r, and rasa formula. 

b. Using your formula, find V when r = 3 ; when r = 6. 

c. How does V change when r is doubled ? 

(Continued on page 32.) 


When r = 

1 

2 

4 

8 

then A = 

? 

f 

? 

f 












32 


ALGEBRA 


12. Each of n bags of grain weighs k pounds. Let w be their 
total weight. 

a. Express in a formula the relation between w, n , and k. 

b. By your formula find w when n — 25 and k = 100. 

c. How does w change when k is doubled ? 

d. How does w change when n is increased ? 

13. a. The postage on parcels going to a certain zone is 10^ 
for the first pound and for each additional pound or fraction 
of a pound. Let p = the postage and (n + 1)= the number of 
pounds weight. What is the formula connecting p and n ? 

b. Find p when n = 3 ; when n = 5 ; when n = 10. 

c. Is p doubled when n is doubled ? 

d. What does happen to p when n increases ? 

14. a. An agent receives a commission of c dollars for each 
article that he sells. If C represents his total commission from 
the sale of n of the articles, express by a formula the relation 
between C; c, and n. 

b. Find C when c = $1.25 and n = 14. 

c. Find C when c = $1.25 and n = 7. 

d. If c has a fixed value, what happens to C when n is halved ? 
When n is trebled ? 

15. a. Another agent receives each day w dollars in wages 
and, besides, a commission of c dollars on each article that he 
sells. Let T represent his total income for a day when he sells 
n articles. Express the relation between T , w, c, and n. 

b. Find T when w = $2.50, c = $.50, and n = 24. 

c. Find T when w = $2.50, c = $1.00, and n = 24. 

d. If w and n have fixed values, what happens to T when c 
increases ? When c is doubled ? 

Mv 2 

16. F = (Formula from physics.) 

Find F when M = 175, v = 25, g = 32, and r = 5. 


II. POSITIVE AND NEGATIVE NUMBERS 


26. Another characteristic of algebra is the use in it of num¬ 
bers called positive and negative numbers. These numbers 
are used to distinguish between opposite quantities. 

Rising and falling temperature are opposite quantities. A 5° rise 
in temperature followed by a 5° fall has a total effect of no change at all. 

Similarly, a $50 gain and a $50 loss are opposite quantities. A gain 
of $50 followed by a loss of $20 has the same effect as a single gain of 
$30. 

EXERCISE 19 

What is the opposite of: 

1. Sailing 25 miles north ? 4. Depositing $20 in a bank ? 

2. Walking 10 steps forward? 5. A 10° fall in temperature? 

3. Gaining 15 pounds in weight? 6. Adding 50? 

What is the total effect of: 

7. Winning 15 points in a game and then losing 10 points? 

8. Walking 12 steps forward and then 8 steps backward? 

9. A 10° rise in temperature followed by a 15° fall? 

10. Walking 20 steps to the right and then 25 steps to the 
left? 

11 . An 8° rise in temperature followed by a 15° fall? 

12. Driving 25 mi. north and then 30 mi. south? 

13. First adding 36 and then subtracting 30? 

14. First adding 18 and then subtracting 10 ? 

15. First adding 18 and then subtracting 25? 

16. Gaining $150 and then losing $80? 

17. Gaining $40 and then losing $75 ? 

33 


34 


ALGEBRA 


27. The signs + and — are used to distinguish between 
opposite quantities. 

8° above zero is written + 8°. 8° below zero is written — 8°. 

A $5 gain may be written + $5; then a $5 loss, — $5. 

A quantity preceded by a plus sign is called a positive quan¬ 
tity ; one preceded by a minus sign is called a negative quantity. 


EXERCISE 20 


1. The temperature at 8 a.m. was + 10°. At noon, it was 
15° warmer. What was the temperature then ? 

2. At 7 a.m., the temperature was + 25°. By 6 p.m., the tem¬ 
perature was 15° lower. What was the temperature at 6 p.m. ? 

3. a. At 7 a.m., the temperature was — 5°. By noon it 
had risen 5°. What was the temperature at noon ? 

b. By 6 p.m., it had fallen 8° from the temperature at noon. 
What was the temperature at 6 p.m. ? 

4. What is the total effect of (+ 15 steps) + (— 10 steps) ? 
Note. — Read this: “what is the total effect of 15 steps to the right 

followed by 10 steps to the left ? ” 

Write the solution as follows : 

(+ 15 steps)+(— 10 steps) = (?) steps. 

Similarly read and write the result of: 

5. (+25 steps) + (— 10 steps) 8. ( — 5 steps) + (+ 8 steps) 

6. (+15 steps) + ( — 20 steps) 9. ( — 8 steps) + (— 2 steps) 

7. (+ 15 steps)+ (— 15 steps) 10. (— 9steps) + (— 4 steps) 

11. (+$15) + (-$20)=? 

Read this : “ a $15 gain followed by a $20 loss gives ” 


Similarly read and give the total result of: 


12 . (+$ 20 )+(+$ 10 )= ? 

13. (+ $12)+(— $5)= ? 

14. (- $8)+(+$ll)= ? 

16. (- $3)+(— $2)= ? 


16. (+$100)+(-$25)= ? 

17. (-$75)+(+$50)= ? 

18. (+ $60)+(— $80)= ? 

19. (- $20)+(- $15)= ? 


POSITIVE AND NEGATIVE NUMBERS 


35 


28. Positive and negative numbers. In algebra, instead of 
an arithmetical number, like 3, there is the number + 3, called 
positive 3, and the number — 3, called negative 3. 

The sign must always be written before a negative number; 
if no sign appears before a number, the number is positive. 

The arithmetical number which remains when the sign of a 
positive or of a negative number is omitted is called the absolute 
value of the number. 

Thus, the absolute value of + 5, or of — 5, is 5. 

29. Geometrical representation of positive and negative 
numbers. 

t —5 -4 -3-2- 1 0 f| -t-2 -t-3 -1-4 + 5 

A 

0 is represented on this line by point A. 

•+■ 1, +2, + 3, etc., are represented by points on one side of 0. 

— 1, — 2, — 3, etc., are represented by points on the other side of 0. 

The graph at the right uses 
this idea. It shows the tern- + 6 ° 
perature from 8 a.m. to 4 p.m. 
on a winter’s day. +3 ° 

+ 2 ° 

Questions. 1. What was +i° 

the temperature at: 8 a.m.? °° 

9 a.m.? 10 a.m.? 12 m.? 

2. How many degrees did -3° 
the temperature rise between 

10 a.m. and 1 p.m. ? 

3. How many degrees did it drop between 3 and 4 p.m. ? 

4. How much warmer was it at 12 m. than at 8 a.m. ? 

6. How much colder was it at 4 p.m. than at 1 p.m. ? 































































36 


ALGEBRA 


30. Addition of positive and negative numbers. 

The fundamental fact about two numbers like (+ 3) and (— 3) 
1S that (+ 3) + (- 3)= 0. 

Similarly (— 5)+(+ 5)= 0; (+ #)+(— x)= 0, etc. 

The rules for adding positive and negative numbers are sug¬ 
gested by the following examples. 

Just as $6 gain followed by $4 gain gives $10 gain, 
so (+ 6) + (+ 4) = + 10. 

Just as $6 loss followed by $4 loss gives $10 loss, 
so (— 6)-f-( — 4) = — 10. 

Just as $6 gain followed by $4 loss gives $2 gain, 
so (+ 6) + (— 4) = + 2. 

Just as $6 loss followed by $4 gain gives $2 loss, 
so (— 6) + (+ 4) = — 2. 


Rule. — 1. To add two positive numbers, add their absolute 
values (§ 28) and prefix the plus sign to the result. 

2. To add two negative numbers, add their absolute values 
and prefix the minus sign to the result. 

3. To add a positive and a negative number, subtract the 
smaller absolute value from the larger and prefix to the result the 
sign of the number having the larger absolute value. 


31. Adding a negative number gives the same result as sub¬ 
tracting the corresponding arithmetical number. 

Thus, 12 +(— 3) gives the same result as 12 — 3. Each equals + 9. 

In arithmetic, the sum of two numbers is always greater than 
either of them. In algebra, the sum of two numbers will be 
smaller than one of them if one is negative. This fact is confusing 
to the beginner. To emphasize it, the sum of two numbers in 
algebra is often called their algebraic sum. 


POSITIVE AND NEGATIVE NUMBERS 


37 


EXERCISE 21 


1. 

Add + 3 to : 

+ 8; +7; 

+ 10; 

+ l; 

+ 3. 


2. 

Add — 5 to: 

-4; - 6; 

- 12; 

- 7; 

- 1. 


3. 

Add — 6 to: 

+ 10; +8: 

; +6; 

+ 3; 

+ 1. 


4. 

Add + 8 to : 




- 2. 


6. 

Add -{- 12 to 

+ 2; -3: 

; - 15; 

+ 4: 

; - 6. 


6. 

Add — 10 to 

+ 5; -4: 

I - 7; 

+ ll; 

; + 3. 


7. 

Add + 7 to: 

— 4; + 1; 

- 11; 

- 7; 

- 2. 


8. 

Add — 9 to: 

— 1; + 1; 

+ 12; 

- 3; 

+ 9. 


Find the following indicated sums : 




9. (+5)+(+4) 

13. (+ 11)+(- 15) 17. 

(— H)+(+ 16) 

10. (- 

- 7)+(— 3) 

14. (+!)+(- !) 

18. 

(+ 14)+(— 6) 

11. (- 

- ll)+(+ 7) 

15. (— 16)+(— 2) 

19. 

(+ 12)+(- 19) 

12. (+9)+(-5) 

16. (-13)+(+9) 

20. 

(- 15)+(— 7) 

21. 

22. 

23. 

24. 


26. 

26. 

- 6 

+ 15 

- 24 

- 35 

+ 28 

- 19 

+ 22 

- 35 

+ 17 

- 6 


37 

+ 26 

27. 

28. 

29. 

30. 


31. 

32. 

+ 42 

- 74 

- 55 

+ 84 


37 

- 19 

- 36 

+ 60 

+ 37 

- 56 

— 

18 

+ 32 

33. 

34. 

36. 

36. 


37. 

38. 

+ 5 

- 6 

+ 12 

- 6 

+ 14 

- 8 

- 7 

- 4 

- 10 

- 11 

+ 13 

+ 11 

- 3 

+ 8 

+ 5 

+ 10 

- 

9 

- 3 

+ 4 

- 2 

- 7 

+ 6 

+ 

6 

+ 5 

39. 

40. 

41. 

42. 


43. 

44. 

+ 9.2 

- 7.3 

- 8.5 

+ 10.6 

— 

2.4 

- 10.5 

- 3.6 

+ 6.2 

- 3.7 

- 7.5 

+ 5.7 

- 6.4 


Find the sum of: 

46. — and + -J 47. — -J and — % 49. + 2\ and — 3^- 
46. -f - -jr and — -g- 48. + and 60. 3^- and + 2^- 





























38 


ALGEBRA 


32. Multiplication of positive and negative numbers. 

The words multiplier, multiplicand, and product have the 
same meaning in algebra as in arithmetic. 

The rules for multiplying positive and negative numbers are 

suggested by the following examples. 

The sign X is to be read “ multiplied by” 

a. (+'4)X(+ 3) — (+ 4) + (+ 4)+(+ 4), or + 12. 

.*. (+ 4)X(+ 3) = + 12. 

b. (- 4)X(+ 3) = (— 4) + (— 4)+(— 4), or - 12. 

/. (- 4)X(+ 3) = - 12. 

c. Just as 4 X 3 = 3 X 4, in arithmetic, so it is assumed that 
(+ 4)X(— 3) should equal (— 3)X(+ 4) in algebra. 

Since (— 3)X(+ 4) = (— 3)+(— 3)+(— 3)+(— 3) or — 12, 

/. also (+ 4)X(- 3) = — 12. 

d. It remains to find (— 4)X(— 3). The fact is that 

(- 4)X(- 8) = + 12. 

Mathematicians accept this fact without any proof. Explana¬ 
tions like the following are often given, but they are not proofs. 
Think of — 4 as losing $4. 

Then (— 4) X (+ 3) would be losing $4 on three occasions or losing 
altogether $12, and this is represented by — $12. 

If (— 4) X (+ 3) means losing $4 on three occasions, 
then (— 4) X (— 3) may be thought of as meaning failing to lose $4 on 3 
occasions. Of course, if a person fails to lose $4 three times, he really 
has $12 more than he might have had. From this (— 4) X( — 3) gives 
( + $12). 

Rule. — To multiply one signed number by another: 

1. Find the product of their absolute values. (See all four 
examples above.) 

2. Make the product positive if the multiplicand and multiplier 
have like signs. (See Examples a and d above.) 

3. Make the product negative if the multiplicand and multi¬ 
plier have unlike signs. (See Examples b and c above.) 


POSITIVE AND NEGATIVE NUMBERS 


39 


EXERCISE 22 

1. Multiply + 3, + 7, + 6, + 11, and + f by + 5. 

2. Multiply + 4, + 2, + 8, + 9, and + -J by — 3. 

3. Multiply — 2, — 5, — 7, — 10, and — f by + 4. 

4. Multiply — 3, — 5, — 6, — 9, and — -J by — 2. 

5. Multiply + 2, — 5, + 3, — 6, and + 7 by + 6. 

6. Multiply — 3, + 4, — 8, + 9, and — f by — 7. 

7. Multiply + 16, — 12, + 10, — 14, and — 20 by — 

8. Multiply — 21, + 15, — 12, — 27, and — 6 by — + 

9. Multiply + & ~ i, ~ f, - i , and - * by + 24. 

10. Multiply ~ + b ~ i, + f, and - A by - 36. 


Find the following indicated products : 


11 . 

12 . 

13. 

14. 


(+ 6)X (— 9) 
(- 7)X(— 11) 
(- 8)X(+ 12) 
(- 10)X(- 14) 


15. 

16. 

17. 

18. 


(+ 9)X 0 
(- 8)X(+ i ) 
0X(- 12) 

(- 1)X(- 5) 


( — i)(+ i) 


(+ i)(- 


19. 

20 . 

21 - (~T 

22 . (-!)(+ 


i) 

L )(- *) 
i) 


23. (-1)X(-1)X(-1) 

24. (- l)x(-2)x(+2) 

25. ( — 3) X( — 3) X( — 3) 


26. ( — 2) X( — 3) X(+ 4) 

27. (-1)X(+4)X(~2) 

28. (-6)X(-2)X(-1) 


REVIEW EXERCISE I 

1. Give the sum and also the product of: 

a. + 12 and — 4; c. — 20 and —5; e. — 7.5 and — .3; 

b. —18 and+6; d. +4.3 and—1.8; /. — fand+-J. 

2. Give an illustration which helps you remember how to 
add + 25 and — 35; also how to add — 15 and — 20. 

3. Eight fifteenths of a certain number is 235. Find the 
number. 

4. Solve the equation 3s — 14s + 22s = 73.7. 

6. The perimeter of a certain triangle is 63 inches. The 
second side is 2 times as long as the first side and the third side 
is 3 times as long. Find the sides of the triangle. 


40 


ALGEBRA 


33. Evaluating expressions for negative values of the literal 
numbers. 

Example. What is the value of ax 3 + bx + c when a = 5, 
b — 2, c = — 3, and x = — 2 ? 

Solution. 1. ax* + bx + c = 5(- 2) 3 + 2(- 2)+(- 3) 

2. =5(—8) + (—4)+(—3) 

3. = - 40 - 4 - 3, or - 47. 

Note. — Observe : that + bx means add b • x; that b • x is 2(— 2), 
or — 4; that + (— 4) gives — 4 in Step 3 by the rule in § 31, page 36. 


EXERCISE 23 


Find the values of the following expressions, when x = 4, 



3, b = - 2, 

c — — 1. 




1. 

6 a 2 x 

8. 

a 2 + x 2 

15. 

z 2 + 2 * + 1 

2. 

3 bx 2 

9. 

a 2 + b 2 

16. 

b 2 + 2 b + 1 

3. 

abcx 

10. 

c 2 + a 2 

17. 

c2 + 2c+ 1 

4. 

— 5 c 2 x 

11. 

ax + be 

18. 

ax + ab + ac 

6. 

— 2 ab 2 c 

12. 

b 2 + a 2 

19. 

a 2 + b 2 - c 2 

6. 

ax 

13. 

a 2 + x 2 

20. 

x 2 - 2 x + 1 

be 

b 2 + c 2 

a 2 - 2 a + 1 

7. 

4 ab 

14. 

a 2 + 9 c 2 

21. 

a 2 — b 2 + c 2 

3 cx 

ax + b 

a 2 — b 2 — c 2 


22. If F = fC + 32, findF: 

(a) When C = — 10; b) when C — — 15; (c) when 

C = - 20. 


23. S = at — % gt 2 . 

Find £ when a = 200; t = 3; gr = 32. 


24. 


ad 2 , 


а. Find p when a=.6; d = \ \ t = -J. 

б. Find p when a = .6; d = \ ) t = i- 

c. Find p when a=.6; d = \ ) t = \- 

d. If a and d have fixed values, what change in p is caused 

by a decrease in t ? 






m. ADDITION AND SUBTRACTION OF 
POLYNOMIALS 


34. A monomial or term consists of numbers connected only 
by signs of multiplication or division. Thus, 

2 x 2 , — 3 ab, and + 5 are the terms of the expression 2 x 2 — 3 ab + 5. 

If the sign before a term is plus, as 2 x 2 , it is called a positive term; 
if minus, as — 3 ab, it is called a negative term. 

35. If two or more numbers are multiplied together, each of 
them or the product of any number of them is a factor of the 
product. 

Thus, a, b, c, ab, ac, and be are all factors of abc. 

36. Any factor of a product is called the coefficient of the 
product of the remaining factors. 

Thus, in the term 2 ab, 2 is the coefficient of ab, 2 a of b, and 2 b of a. 

In a term like 5 x?y, the factor 5 is called the numerical 
coefficient of x 2 y. 

A term like x has the numerical coefficient 1, since x is the same as 1 x. 

If a term is negative, like —3 a, the numerical coefficient is — 3, 
since (— 3) • a = — 3 a. 

37. Like terms are terms which have the same literal factors. 

Thus, 2 x 2 y and — 5 x 2 y are like terms; also, 2 and 3 

Unlike terms are terms which do not have the same literal 
factors. 

Thus, 2 x 2 y and 2 xy 2 are unlike terms. 

Unlike terms may be like with respect to one or more letters. 

Thus, 2 axy and 3 byx are like with respect to xy. 

41 


42 


ALGEBRA 


38. Addition of like terms with positive coefficients has been 
taught previously. Thus, 6n + 4w = 10 n. 

n is the common factor of 6 n and 4 n. 6 and 4 are the coefficients 
of n. Their sum, 10, is the coefficient of the result. 

Rule. — To add two or more like terms, multiply their common 
factor by the sum of its coefficients. 

Example. What is the sum of — 5 x 2 and + 3 x 2 ? 

Solution. 1. These are like terms, having the common factor x 2 . 
The coefficients of x 2 are — 5 and + 3. 

2. The sum of — 5 and + 3 is — 2. 

3. .*. (- 5 x 2 ) + (+ 3 x 2 ) = — 2 x 2 . 

Note. — If x = 2: (- 5 x 2 ) = - 20; + 3x 2 =+12; 

and — 2 x 2 = — 8. 

Also (- 20) + (+ 12) = - 8. 

No matter what the value of x, (— 5 x 2 ) + (+ 3 x 2 ) = — 2 x 2 . 


EXERCISE 24, a 


Find the following sums : 


1. 

2. 

3. 

4. 

5. 

— 9 p 

+ 6 r 

- 5 xy 

— 7 rs 

+ 

00 

+ 11 P 

- 10 r 

+ 12 xy 

— 6 rs 

- 14 x 2 

6. 

7. 

8. 

9. 

10. 

- 5 t 

+ 17 a: 2 

— 10 ab 

+ 6 xyz 

- ab 2 

- 9 t 

- 12 a: 2 

— 13 ab 

— 13 xyz 

- 8 ab 2 

11. 

12. 

13. 

14. 

15. 

+ 3 c 

+ ix z 

- 2 yz 

+ 4 xy 2 

— 3 abc 

-5c 

- z 3 

+ 

1 

00 

to 

— 5 abc 

+ 6c 

-7x> 

~ 5 yz 

+ xy 2 

+ 2 abc 


16. 

(+8^)+(- 15 A) 

20. 

17. 

(+9p)+(- 12 p) 

21. 

18. 

(— 11 m)+(+ 8 m) 

22. 

19. 

(- 9 x 2 )+(+ 2 x 2 ) 

23. 


(— ab 2 c) + (— 8 ab 2 c ) + (+ 14 ab 2 c) 
(+6xy) + (- 13a:y) + (+ 5 xy) 

( — 10 a&)+(+ 5 ab)-\-{ — 13 ab ) 
(— 9 i 2 )+(— 4 £ 2 ) + (+ 6 t 2 ) 

















ADDITION AND SUBTRACTION OP POLYNOMIALS 43 


24. 

25. 

26. 

27. 

28. 

— 5 t 

+ 17 x 2 

+ 9 ab 

+ xyz 

+ 7 ab 2 c 

+ 12 t 

CO 

1 

— 10 

— 4 xyz 

— ab 2 c 

- 9 t 

<M 

00 

+ 

— 13 ab 

+ 6 xyz 

— 8 ab 2 c 

- 4 t 

- 12 X 2 

+ 7 ab 

— 13 xyz 

+ 14 ab 2 c 

+ 6 t 

+ 5 x 2 

-f- 5 ab 

+ 5 xyz 

— 9 ab 2 c 

29. 

30. 

31. 

32. 

33 . 

— mn 

- f rs 

_ 3 -.3 

5 X 

— 2.5 ab 

- .25 y 2 

+ ^ mn 

+ irs 

+ ?x* 

+ 3.4 ab 

-3.5 j/ 2 

— mn 

— in 

_ 4 ^.3 

10 ^ 

— .6 ab 

+ 4.5 y 2 


EXERCISE 24 b 

Note. — Terms having literal exponents. In § 25, you learned that 
x 3 means x • x • x; that x 9 means that x is used as a factor nine times. 
For the present, the symbol x n has meaning only if n represents some 
positive integer (like 11 or 26). Its meaning when n is a negative integer 
or a fraction will be given later. x n is read, “z to the power n.” 

When terms like 3 x n and — 7 x n appear in the same example, it is 
understood that n represents the same positive integer. These terms 
are like terms and can be added, giving — 4z n . 

Similarly, (+ 6 x a y h ) + (— 2 x a y b ) = + 4 x a y b . 


1. What is the numerical value of x n , when x = 3, and : 
a. n — 2? b. n = 3? c. 

Find the following sums : 


2 . 

— 4 x m 
-f- 9 x m 

7. 

_ 2 /y.n+1 

-3 X 

+ i 


3. 

+ 8 x n y 2 

— 10 x n y 2 

8 . 

— | x a y n 

— j x a y n 


4. 

— 3 x a y b 

-|-ll x a y b 

9 . 

_ 3 ^.m 

■5 xy 

+ r 3 o x y m 


n —— 4 ? 

5. 

— 14 mfx a 

— 6 m s x a 

10 . 

- 6.05' x 2n 

- 3.75 x 2n 


d. n = 5? 


6 . 

+ 8.5 z n w m 

— 4.2 z n w m 

11 . 

- 8.27 z 3a 
+ 2.17 z 3a 






















44 


ALGEBRA 


39. Addition of unlike terms can only be indicated, and 
that is done by writing them with a plus sign between them. 

If asked to add 5 gal. and 3 qt., you can write 5 gal. + 3 qt. 
This expression indicates the sum. You get the actual sum 
only when you change 5 gal. to 20 qt. Then, 5 gal. + 3 qt. = 
20 qt. + 3 qt., or 23 qt. 

Similarly, if asked to add 5 x and 3 y, you can only indicate 
the sum by writing 5 x + 3 y. You cannot combine 5 x and 
3 y in any way since they are unlike terms. Later, if you learn 
that x = 4 and y = 6, you can find the value of the sum. Then 
5 x + 3 y = 20 + 18, or 38. 

Example. Write in simplest form the sum of 2 x 2 , — 3 x, 
— 4, — 5 x 2 , + 7, and + 8 x. 

Solution . 1. Indicating the addition, the sum is 

(2 x 2 ) + (- 3 *) + (- 4) + (- 5 z 2 ) + (+ 7) + (+ 8 *). 

2. This equals 2 z 2 - 3 x - 4 - 5 z 2 + 7 + 8 x. (§31) 

3. Combining like terms, the sum is — 3 r 2 + 5 a: + 3. 


EXERCISE 25 


Indicate the sum of the terms in each of the following ex¬ 
amples, and simplify the result by combining like terms: 


1. 7 ab, — 3 ac, -f 4 ab 

2. 9 r 2 , — 7 r, + 4 r 2 , — 8 r 

3. 12 xy, — 10 xy , + 6 xz 

4. — 4 z, + 7 z, — 10 y 8. — fw 

9. 8 x 2 , — 9 x, — 6 x 2 , + 5, + 4 x, — 7 


5. 16 t 2 , - 1 It 2 , - 12 t 
6- - i xy, %xy, - f xz 
7. |m 2 , - i m 2 , - T \ 7 
» 2 y + tv v> 


i 

f T 


10. 4 a, —5 b, — 3 c, + 7 b, + c, —2a 

11. 10 m 3 , -f- 6 n 2 , — 8 mn, — 4 m 3 , + 5 mn, and — 2 n 2 

12. 10 r 2 , —6 r, + 5 s, — 8 r, +2 s, — 5 r 2 , and —4 s 

13. 9 r 3 , - 6 r 3 , + 3 r 2 , - 7 r, + 8 r 2 , - 15 r, and + 5 

14. 2.5 x 2 , — 1.2 xy, + .5 xy, — 3.8 x 2 , and + 2 y 2 

15. 2.4 a 3 , -5a 2 , + .7 a 3 , - 3 : + 2 a 2 , and - 1 


ADDITION AND SUBTRACTION OF POLYNOMIALS 45 


40. A polynomial is an algebraic expression of two or more 
terms, as, a + b or x 2 — 2 xy + 5 y 2 . 

A binomial is a polynomial having two terms; as a + b. 

A trinomial is a polynomial having three terms; as x 2 — 2 xy 
+ 5 y\ 


41. Arrangement of terms in either ascending or descending 
powers of a letter. 

If the terms of a polynomial are — 3 x 2 y, + % 3 , — y 3 , and 
+ 3 xy 2 , a natural orderly arrangement of them is as follows: 
x s — 3 x 2 y + 3 xy 2 — y 3 . 

These terms are arranged in descending powers of x. 

The exponents of x in order are now 3, 2, 1, and 0, because 
— y 3 does not have any factor x. 

The terms are arranged in ascending powers of y. 

The exponents of y in order are 0, 1, 2, 3, for the first term, 
x 3 , does not contain any factor y. 

42. Addition of polynomials. 

Example. Add 4a+56 — 2c and — 3 a 2 b — 6 c. 

Solution. 4a+55 — 2c 

— 3 a + 2 5 — 6c 
a + 7b - 8c 


In this example, a, b, and c may represent any numbers. 


Thus, if a = 2, b = 3, and c = 1, 

4a + 5& — 2c = 8 + 15 — 2 =21 
— 3a + 26 — 6c = — 6 + 6 — 6 = —6 


21 +(- 6) = 15. 


Also, the sum a + 76 — 8c = 2+21 —8 = 15. 

This fact is most important. It can be used when checking a solution. 

Rule. — To add polynomials: 

1. Rewrite the polynomials, if necessary, in descending 
powers, or ascending powers of one letter, with like terms in 
the same column. 

2. Add the columns of like terms and combine the results. 

3. Check the solution by adding again in the opposite direc¬ 
tion, or by substituting values for the letters. 



46 


ALGEBRA 


EXERCISE 26 

Find the following sums, checking each hy adding in the oppo¬ 
site direction: 


1. 

+ 7 z * 2 - 2 a; + 1 
-2z 2 + 4z- 3 
— 3 x 2 -f- x -f- 2 

4. 

a 2 - 3 a + 1 
a 2 + 9 a — 6 

2 a 2 +3 


2 . 

- 10 A + 5 B 
+ 44-35 
+ 2 A - 6 B 

5 . 

x 2 + 2/ 2 

2 a; 2 + xy 

— 3 xy — 2 y 2 


3. 

+ 3m-4w+7y 
-5m + 9n-6p 
+ 7to - 6w-5|j 

6 . 

m 2 + 2 mn + n 2 
2 m 2 —3 n 2 

— 3 mn 2 n 2 


Note. — Observe that there are blank spaces in Examples 4, 5, and 
6 for powers which are missing from the polynomials. 

Illustrative Example. Find the sum of x 3 — y s + 3 xy 2 — 3 x 2 y, 
2 x 3 + 3 xy 2 + y 3 , and 2 x 2 y — 4 xy 2 . Check, by letting x = 2 
and y — 1. 

Solution. 1. Arrange the terms in descending powers of x , with 
like terms in the same column. 

x 3 - 3 x*y + 3 xy 2 - y* = 8 - 12 + 6 - 1 = 1 
2x z + 3 xy* + y 3 = 16 + 6 + 1 = 23 

+ 2 x 2 y — 4 xy 2 = 8 — 8 =0 

2 x 3 — z 2 ?/ + 2 a:?/ 2 24 

2. Also, 3 x 3 — x 2 y + 2 xy 2 = 24 — 4 + 4 = 24. 

3. .*. The solution is correct. 

Arrange the following examples as in Examples 1-6 and find 
the sums. Check 7-11 hy substitution; the remainder hy reverse 
addition. 

7. 6a—56+7c and 4a + 66 — 4c 

8. 5k — 7 l + 3 m, 4 & + 3 / — 5 m, and 8 k — 2 l 

9. 3a + 9& — 6 c, — 46 + 3c—6d, and 7c—3a + 4d 

10. x 2 — 2 xy — y 2 , x 2 + 2 xy — y 2 , and 2 x 2 + 2 y 2 — 5 xy 

11. 5 m 2 — 3 mn + n 2 , m 2 — 4 mn — 4 n 2 , and — 3 m 2 + 6 n 2 

12. a 3 + 3 a 2 6 + 3 ah 2 -f~ h 3 and a 3 — 3 a 2 6 — b 3 + 3 a6 2 

13. 3 x 2 — 8 xy + 4 y 2 , 2 xy — 4 £ 2 — 7 y 2 , and 6 y 2 + 3 a:?/ 









ADDITION AND SUBTRACTION OF POLYNOMIALS 47 


14. 15 a 3 — 7 — 3 a + 16 a 2 and + 4 + 9 a — 20 a 2 — 10 a 3 

15. — 4 x + 2 x s — 5, 6 x 2 + 6 x 3 , and 13 — 2 x 3 + 9 x 

16. 2 a 2 - 3 6 2 , — 3 a6 + 8 6 2 — 5 a 2 , and 9 a 2 + 6 ab - 10 6 2 

n. i a — ±b + ± c and — | a + -J 6 — c 

18. f m + f n — r, — m — t 7 q w r, and m + f r — -J w 

19. 2.5 ^ - 6.25 B + 4.8 C and - 1.8 A + 5 B - 3.4 C 

20. 2.25 x 2 - 3.4 xy - 2 y 2 and - 1.75 z 2 + 2.1 xy - 1.5 y 2 

REVIEW EXERCISE II 

1. What is : a. a monomial ? 6. a polynomial ? 

2. What are like terms? What are unlike terms? 

3. What does x 5 mean ? What is its value when x = — 2 ? 

4. What is the value of x a y b when x = 2, a = 3, y — — 2, 
and 6 = 4? 

5. Which of the following statements are true and which 
false: 

a. (- 4)+(- 3) = + 7? 6. *(- 2) • (- 3) = - 6? 

c. (— 5)+(+ 8) = + 3? d. (+ 4) • (— 2) = — 6? 

6. Solve the equation lOw — 4:W-{-3w — w = 96. 

7. If f x — 24, is it true or false that x = 32 ? 

8. By the formula V = r 2 h, find h when V = 4312, r = 7, 
and 7r = -^r 2 -. 

9. Find the sum of 3 x 2 — 5 + 8 x 3 — 7 x; 12 — 5 x 2 -f- x s 

+ 9 x; and 6a:-4x 2 + 2x 3 - 11. Check by letting x = 2. 

10. Find the sum of 3 y 2n — 4 y n + 5; 7 + 5 y 2n + 6 y n ; 

and y n — 4 y 2n — 11. 

11. Complete the following sentences : 

а. The sum of two negative numbers is a_number. 

б. The product of two negative numbers is a_number. 

12. If 7 apples cost 25^, what will x apples cost? 

13. If n apples cost 50^, what will y apples cost? 


48 


ALGEBRA 


43. Addition used in solving equations. 

In the adjoining figure, suppose the sugar S just balances the weight 
W. If a 5 lb. weight is added to the right scale pan and 5 lb. of sugar 
to the left pan, then the two pans will still balance. 

Also the two pans A will not balance if a 5 lb. 
weight is added to the right pan and 3 lb. or any¬ 
thing other than 5 lb. to the left. 

Similarly, if 3 x = 60, then 3 x + 5 = 60 + 5. 

Rule II. The same number can be added to both sides of an 
equation without destroying the equality. 

Example 1. Solve the equation x —5 = 13. 

Solution. 1. x — 5 is 5 less than x. If 5 is added to x — 5, the 
result is x — 5 + 5 or x. 

But, if 5 is added to the left side of the equation, it must also be 
added to the right side in order to keep the two sides equal. 

Therefore, add 5 to both sides of the equation. 

2. Then, Z — 5 + 5= 13+5. (Rule, above.) 

3. Combining like terms, x = 18. 

Check. Substituting 18 for x in the original equation, 

does 18 - 5 = 13? Yes. 

Then, 18 is the root of the equation. 

Hereafter, instead of writing in full 
“ add 5 to both sides of the previous equation we shall write As, 
and instead of “ combining like terms ” we shall write C. T. 

Example 2. Solve the equation 5 y — 13 = 22. 

Solution. 1. 5 y — 13 = 22. 

2. Ais 5 y = 35. 

3. Ds y = 7. 

Note. — In Step 2, what does Ai 3 mean ? Observe that 5 y —13 + 13 
gives 5 y; that 22 + 13 = 35. This should be done mentally. 

Check. Substitute 7 for y in the original equation. 

Does 5 • 7 - 13 = 22? 

Does 35 - 13 = 22? 

Yes. Therefore 7 is the root of the equation. 









ADDITION AND SUBTRACTION OF POLYNOMIALS 49 


EXERCISE 27 

Solve and check the following equations : 


1. x - 10 = 5 

2. y - 12 = 16 

3. 2z — 5 = 13 

4. St — 7 = 14 
6. 4 n — 9 = 27 

11 . Solve the equation 7 x 


6. m — 14 = - 8 

7. ^ - 16 = - 3 

8. 2 B - 21 = - 5 

9. 5 t - 28 = - 8 
10 . 6 r - 40 = - 2 

55 — 4 a\ 


Solution. 1. 7 x = 55 — 4: x. 

We must remove 4 x from the right side. 
2. A^ 11 z = 55. 

(Complete the solution.) 

Note. —What does A 4a; in Step 2 mean? 


7 x + 4x = 11 a? 
55 — 4x 4x = 55 
(Do this mentally.) 


12. 5 x = 32 — 3 x 15. 8 w = 130 — 5 w 

13. 7 t = 27 - 2 t 16. 16 A = 13 - 10 A 

14. 3 y = 42 — 4 y 17. 25 v = 8 — 15 v 

18. Solve the equation 3 x — 42 = — 4 x. 


Solution. 1. 3 x — 42 = — 4 z. 

We must remove — 42 from the left side. 
2. A 42 3 x = 42 — 4 x. 

(Complete the solution.) 


3z — 42 +42 = 3z 
— 4 x + 42 = 42 — 4 x 
(Do this mentally.) 


19. 

5 y - 

- 70 = 

= -2y 

28. 

3c — 3 = 13—c 

20. 

7 m 

- 80 

= — 3 m 

29. 

II 

rH 

1 

^3 

1 

CO 

1 

21. 

12 s 

- 95 

= - 7 s 

30. 

- 16 + 2 a: 

= 9 - Sx 

22. 

18 x 

- 24 

= — 6 x 

31. 

7 w - 18 = 

6 — w 

23. 

9 z - 

- 4 = 

- 3 z 

32. 

13 5 — 11 = 

■ 4- 2s 

24. 

20 r 

- 24 

= - 12 r 

33. 

5 2/ — 8 = - 

-57/ -3 

25. 

5c - 

- 2 = 

23 

34. 

6 717 — 11 = 

88—3 w 

26. 

15 x 

= 45 

- 15 x 

35. 

II 

1 

I- 

rH 

1-71 

27. 

4b : 

= 20 

- 6 

36. 

c — 6 = 14 

— c 




50 


ALGEBRA 


44. Subtraction of positive and negative numbers. 

a. The meaning of subtraction. To subtract a from b means 
to determine the number which must be added to a to give b. 

Thus, to subtract — 3 from + 8 means: “What number must be 
added to — 3 to give +8?” The answer is + 11. 

On the scale below, to go from — 3 to + 8, you must count 11 spaces 
to the right. 

t: ~i ~ 3 ~1 o -H 4-2 4-3 +4 +5 ^ 

A 

b. The number subtracted is called the subtrahend. The 
number from which the subtrahend is subtracted is called the 
minuend. The result is called the difference or remainder. 

EXERCISE 28 

What number must be added to the first number of each of 
the following examples to get the second ? 

1. 3, 10 3. 10, 19 5. 4 m, 8m 7. 4 c, 26 c 

2. 8, 15 4. 3 a, 5 a 6.9 t, 17 t 8. 2 b, 23 b 

On the scale above, in what direction and how many spaces 
must you count to pass from : 

9. +1 to +4 11. -3 to +2 13. -5 to +8 16. +10 to -5 
10. +5 to -4 12. -1 to -5 14. -6 to -2 16. -4 to -12 


What number must be added to the first number of each of 
the following examples to get the second ? 


17. 

+ 

4, 

+ 

10 

24. 

+ 

10, - 5 

31. 

- 

15, 


- 10 

18. 

- 

5, 

0 


26. 

- 

10, +2 

32. 

+ 

10, 


- 4 

19. 

- 

4, 

+ 

2 

26. 

- 

2, - 10 

33. 

- 

9, 

+ 

6 

20. 

- 

6, 

+ 

3 

27. 

- 

6,-3 

34. 

+ 

15, 


- 10 

21. 

+ 

5, 

+ 

3 

28. 

- 

4, +5 

35. 

- 

8, 

- 

15 

22. 

+ 

7, 

0 


29. 

- 

7, +8 

36. 

+ 

5, 

+ 

17 

23. 

+ 

8, 

- 

4 

30. 

+ 

3,-6 

37. 

+ 

3, 

- 

12 



ADDITION AND SUBTRACTION OF POLYNOMIALS 51 


45. Rule for subtracting positive and negative numbers. The 

correct result for Example 24 is — 15, since ( + 10) + (— 15) is 
— 5. Observe that you can get this result, (— 15), by changing 
+ 10 to — 10, and adding — 10 to — 5. In Example 31, the 
correct result is + 5. You can get this result by changing — 15 
to + 15, and adding + 15 to — 10. Similarly, in each of the 
foregoing examples, you can get the correct result by use of the 
following rule : 

Rule. — To subtract one number from another, change the 
sign of the subtrahend and add the resulting number to the 
minuend. 

Example. Subtract — 6 from — 13. 

Solution. Mentally change — 6 to + 6; then add + 6 
and — 13. The result is — 7. 

Check. The sum of — 6 and — 7 is — 13, as it should be 

EXERCISE 29 


Subtract the lower number from the upper number: 


1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

+ 16 

+ 20 

+ 12 

+ 9 

- 15 

- 13 

- 5 

- 8 

+ 7 

+ 13 

- 4 

- 5 

+ 8 

+ 4 

+ 7 

+ 10 

9. 

10. 

ii. 

12. 

13. 

14. 

15. 

16. 

+ 8 a 

- 7 x 

+ 151/ 2 

- 13 1 

— 19 m 

+ 18+ — 20 x 2 

+ 14 rs 

— 3 a 

+ 6x 

-7 y 2 

— 5 1 

— 6m 

-5 A + 3 x 2 

— 8rs 


17. a. Was it possible to subtract 15 from 12 in arithmetic ? 
h. What is the result in algebra when you subtract 15 from 12 ? 

18. a. Which is the subtrahend in a subtraction example? 
Which is the minuend? 

h. Is it necessary to write the subtrahend below the minuend ? 
19-34. In Examples 1-16, subtract the upper number from 
the lower without rewriting the numbers. 


















52 


ALGEBRA 


46. Subtraction of polynomials. 

Rule. — To subtract one polynomial from another: 

1. Rewrite the minuend, if necessary, in descending powers 
of one letter. 

2. Below it, write the subtrahend, so that like terms are in 
the same column. 

3. Imagine the signs of the terms of the subtrahend changed, 
and add the resulting terms to those of the minuend. 

Example. Subtract 7 ab 2 — 9 a 2 b + 8 6 3 from — 2 a 2 b -f- 
4 ab 2 + 5 a 3 . 

Solution. Arrange the terms in descending powers of a, with 
like terms in the same column. 

Minuend 5 a 3 — 2 a 2 b + 4 ab 2 

Subtrahend — 9 a 2 b + 7 ab 2 + 8 6 3 

Change signs mentally, and add 5 a 8 + 7 a?b — 3 ab 2 — 8 b 3 

Note. — In the first column, since nothing is added to 5 a 3 , the 
result is 5 a 3 . 

In the fourth column, + 8 b 3 becomes — 8 6 3 . Since nothing is 
added to it, the result is — 8 6 3 . 

Check. The sum of the remainder and subtrahend is the minuend. 


Subtract: 

1 . 

5a - 86 + 11c 
2 a + 3 6 — 5c 

4. 

12 a 2 - 10 a - 6 
9 a 2 - 6 a + 5 

7. 

12 r 2 —6 s 2 

9 r 2 — 4 rs + 2 s 2 


EXERCISE 30 

2 . 

11 x -f 13 y — 14^ 
4x + 9y — 16z 

5. 

4x 2 + 13x - 10 
— 3 a: 2 — x — 5 

8 . 


3. 

2r + 7s — 10* 

2r -4s + 2t 

6 . 

3 x 2 — 5 xy + y 2 
2 x 2 — 3 xy +5 y 2 

9. 


a 3 — 3 a 2 6 
a 3 + 3 a 2 6 + 3 ab 2 + 


a: 4 — x 2 y 2 + 












ADDITION AND SUBTRACTION OF POLYNOMIALS 53 


10 . 

x 3 — 5 x 2 +- 6 x 

2x 2 -3x -5 


11 . 12 . 

y z +3 y-2 a 3 - 3 a 2 +4 
— y 2 + 4 y — 1 +- a 2 -- 2 a — 2 


Subtract: 

13. 5a — 6& + 7c from 13 a — 4 6 + 8 c 

14. 2 x 2 — 3 xy + 4 ?/ 2 from 3 a: 2 — 5 xy — 6 y 2 

15. x 2 +- 4 xy —3 y 2 from 2 x 2 — 5 xy +- y 2 

16. 4 z 3 — 6 z 2 + 8 x — 9 from 7 z 3 - 16 x + 18 

17. ax — bx +- cz from 2 ax — bx + cx 

18. — 3 z + 11 x — 4 y from \bx—by + %z 

19. 64-55+30 from 8.4+2R-7C 

20. 2 z 3 — 7 +- 4 x — 5 x 2 from 6 x 2 — 8 + 10 z 3 — 3 x 

21 . From 1 + a 3 - 2 a + a 2 take 3 a — 2 + a 2 — a 3 

22. Subtract 3 x* — 5 + 4 x — 2 z 2 from 7 a; 3 — 5 £ +• 12 — 5 z 2 

23. Subtract 2z 3 — 3z 2 + 4:r — 5 from 0 

24. Subtract from |m - |n+{p 

25. Subtract .5 x — .3 y + 1.4 z from 2.8 x — .5 y + 2.5 z 

26. From the sum of 3 a 2 — ab — 2 b 2 and 2 a 2 + 5 ab — 3 b 2 
subtract a 2 — 3 ab — 4 b 2 

27. From 6 x 2 — 7 xy +- 8 y 2 subtract the sum of x 2 — 3 xy 
— y 2 and 2 x 2 + 5 xy —4 y 2 

28. From the sum of a 2 — 2 ab +- b 2 and 2 a 2 -+ 2 ab + b 2 
take the sum of a 2 — b 2 and a 2 + ab + 3 b 2 

29. How much greater than m 2 + 3 mn — n 2 is 4 m 2 — 5 mn 
+ 6n 2 ? What is the result when m = 3, and n = 2? 

30. How much less than 3z 2 — 7a;+-9is2;r 2 + 4a; — 8? 

31. Subtract 4x 2n - 2x n - 5 from 7 x 2n + 3 x n - 8 


32. Subtract 2 x a - 3 y b + 4 z c from 2.5 a:° - 4.2 y h + 3.5 z c 

33. Subtract i a x + b» c z from 2 a x - | b y + | c z 


Note. — Exercise 157, page 280, can be done now. 





54 


ALGEBRA 


47. Subtraction used in solving equations. 

After reading again the rules in § 13 and § 43, you easily 
understand the following rule. 

Rule III. The same number can be subtracted from both 
members of an equation without destroying the equality. 


Example 1. Solve the equation 2 x + 17 =33. 


Solution. 1. We can remove 17 from the left side of this equation 
by subtracting 17 from it. We must then subtract 17 from the other 
side also to keep the two sides equal. 


2. S w 

3. D 2 

Check. 

Does 


2 x = 16 
x = 8 


^ x 1/ — J 

33 - 17 = 16 
(Do this mentally.) 
Substitute 8 for x in the original equation. 

2-8 + 17 = 33? Does 16 + 17 = 33? Yes. 




Note. — In Step 2, Si 7 means subtract 17 from both sides of the 
preceding equation. 


Example 2. Solve the equation 5f + 6= 22 + 27. 
Solution. 1. 5 t + 6 = 2 t + 27. 

2. S 6 5 t = 2 t + 21. 

3. S 2 < 3* = 21. 

4. D 3 t = 7. 

Check. Substitute 7 for t in the original equation. 


Note. —In Step 2, 5 £ + 6 - 6 = 5 2 < + 27 - 6 = 2 < + 21. 

In Step 3, 5 t — 2 t = 3 t; 2 t + 21 — 2 t = 21. Do this mentally. 


EXERCISE 31 

Solve and check the following equations : 


1. 

a: + 13 = 25 

8. 

10 

x + 5 = 

= 13 + 6 x 

2. 

3 y + 11 = 32 

9. 

13 

y + 3 = 

= 19 + 8y 

3. 

5 t + 7 = 52 

10. 

75 

+ 8 = 

14 + 65 

4. 

4 k = 60 + h 

11. 

15 

m + 7 ^ 

= 7 m + 55 

6. 

7 B = 28 + 3 B 

12. 

10 

t + 9 = 

= 4 1 + 12 

6. 

9 r = 55 + 4 r 

13. 

20 

w + 11 

= 15 w + 12 

7. 

12 2 = 84 + 5 s 

14. 

18 

x + 5 = 

= 15 a; + 7 



ADDITION AND SUBTRACTION OF POLYNOMIALS 55 


Solve the following equations, 
and § 47 : 

using the rules taught in § 43 

15. 

18j/ — 7 = 3y + 23 

23. 

156 + 4 = 11 - 6b 

16. 

3s —3 = 2 + 2s 

24. 

32 IF—6 = 1 + 41U 

17. 

4 2 / + 2 = 3 2 /-J-8 

25. 

10 z — 1 = 6#+.6 

18. 

3 t + 9 = 37 — t 

26. 

8 y + 2 = 3 y + 5.5 

19. 

5 w + 1 = 29 — 2 w 

27. 

7r + 6 = 8.2 — 4r 

20. 

16 r - 13 = 4 r + 23 

28. 

4 c + 6 = 12 — c 

21. 

12a: + 5= 12 — 2x 

29. 

17 t - 4 = 8 - 7t 

22. 

15 6-4 = 3- 66 

30. 

13p + 5= -2p+10 

31. 

Is 3 a root of the equation 6 x 

-7=11 + *? Is-2? 

32. Does — 4 satisfy the equation 

Solve the following equations : 

5 y — 9 = 3y + 6? 

33. 

6z-3.2 = 4- 3z 

38. 

5w)-8 = 3w-f.4 

34. 

8y + 4 = 5y + 7.6 

39. 

9 r — 2.1 = 6 r + 3.6 

35. 

5 z — 2.5 = 2z + 8.6 

40. 

7H + 31 = 3 + + 10.9 

36. 

12 t - 25.6 = 7 t + .9 

41. 

4.7*- 7.2 =1.5*+ 12 

37. 

3.5 x- 15 = 2.1 x - 1. 

42. 

3.2 c - 37 = 1.7 c - 14.5 

48. 

Forming algebraic expressions 

for number relations. 


In arithmetic, you can actually find the sum of two numbers 
like 5 and 8. In algebra, you can only represent the sum of 6 
and x by writing 5 + x. 


EXERCISE 32 

1. Represent 8 more than y ; 10 less than y. 

2. a. How much is 10 increased by 3? 

b. Represent 10 increased by n ; also x increased by 15. 

3. a. How much is 20 decreased by 9 ? 24 decreased by 18 ? 

b. Represent 15 decreased by x ; by m. 

c. Represent y decreased by 4; by 7. 

d. Represent z increased by a. 


56 


ALGEBRA 


4. a. Represent m increased by 6. 

6. What is the value of the expression found in part a , when 
mis 4? 7? 10? -5? -9? 

6. A man spends y dollars out of his income of $200. 

o. How much has he left ? 

b. What is the value of the expression in part a, when y is 
25? 50? 100? 

6. a. How much does 12 exceed 9 ? 22 exceed 15 ? 

b. How much does 5 x exceed 3 x ? exceed 10 ? 

c. How much does y exceed 14 ? 20 exceed x ? 

7. The altitude of a certain rectangle is h inches. Its base 
is 4 in. longer. 

a. Represent its base. 

b. Represent its perimeter. 

8. One number is 4 times as large as a certain smaller num¬ 
ber. The smaller number is represented by s. 

a. Represent the larger number. 

b. Represent 15 more than the smaller number. 

c. Suppose the result’s of parts a and b are equal. Give the 
equation expressing this fact. 

9. One number is 6 times as large as a certain smaller num¬ 
ber. Let x represent the smaller number. 

a. Represent the larger number. 

b. Represent 20 more than the smaller number. 

c. Represent 5 less than the larger number. 

d. Suppose the results of parts b and c are equal. Give the 
equation expressing this fact. 

10 . One number is 13 less than a certain larger number. Let 
g represent the larger number. 

a. Represent the smaller number. 

b. Represent the sum of the two numbers. 

c. Suppose that the sum of the two numbers (represented in 
part b) is 88. Give the equation expressing this fact. 


ADDITION AND SUBTRACTION OF POLYNOMIALS 57 


49. Solving problems by means of equations. 

A problem asks for values of one or more unknown numbers. 

Rule. — 1. Represent one unknown number by a letter. 

2. Represent the other unknown numbers of the problem by 
means of this same letter, using relations of these unknowns to 
the first unknown. 

3. Form an equation expressing an equality which is stated in 
the problem. 

a. Sometimes this equation expresses the fact that certain two re¬ 
sults are equal. 

b. Sometimes this equation expresses the fact that some expression 
containing the unknown number has a definite numerical relation to 
some known number. 

4. Solve the equation, thus getting the unknown number 
represented by the letter. Then find all the other unknown 
numbers required. 

5. Check the solution by determining whether the numbers 
found satisfy the relations given in the problem itself. 

Example. One number is 9 times as large as another. If 
16 be subtracted from the larger, the result is 32 more than the 
smaller. What are the numbers ? 

Solution. 1. Let s = the smaller number. 

2. 9 s = the larger number. 

3. 9 s — 16 = 16 less than the larger number. 

4. s -j- 32 = 32 more than the smaller number. 

5. .*. 9 s — 16 = s + 32, because the problem states 

these two results are equal. 

6 . Ai6 9s = s + 48. 

7. S s 8 s = 48. 

8. D 8 s = 6, the smaller number. 

9. .*. 9 s = 54, the larger number. 

Check. If 16 be subtracted from 54, the result is 38. If 32 be 
added to 6, the result is 38. Since these two results are equal, and since 
54 is nine times 6, then 54 and 6 must be the correct numbers. 


58 


ALGEBRA 


EXERCISE 33 

1. What number increased by 45 equals 63 ? 

2 . One number is 9 times as large as a certain smaller num¬ 
ber. If the smaller number is subtracted from the larger, the 
remainder is 32. Find the number. 

3. The result obtained by subtracting 24 from 7 times a 
certain number is 25. What is the number ? 

4. If 12 be subtracted from 9 times a certain number, the 
result equals 7 times the number. What is the number ? 

6. There are two numbers of which the larger is 7 times the 
smaller. Also the larger is 48 more than the smaller. What are 
the numbers ? 

6. One of two line segments is 4 times as long as the other. 
If 20 in. be added to the shorter, and 4 in. be taken from the 
longer, the two results will be equal. How long are the segments ? 

7. If 13 times a certain number be increased by 5, the result 
is 40 more than 8 times the number. What is the number ? 

8. If 3 be subtracted from 32 times a certain number, the 
result equals 15 more than 20 times the number. What is the 
number ? 

9. Three times a certain number equals 15 diminished by 
the number. 

10. Twice a certain number increased by 5 equals 18 dimin¬ 
ished by three times the number. 

11 . If 9 be added to four times a certain number, the result 
equals 15 more than 3 times the number. 

12. Eight tenths of a certain number increased by 16 equals 
the given number less 10. (Write eight tenths decimally.) 

13. If 9 be added to 7 times a certain number, the result 
equals 16 diminished by the number. 

14. The length of certain lots in one city is 5 times their 
width. The perimeter of these lots is 300 ft. What are the 
length and the width of the lots ? 


ADDITION AND SUBTRACTION OF POLYNOMIALS 59 


50. Problems solved by use of formulas. 

EXERCISE 34 

1. If V represents the perimeter of a triangle whose sides 
have lengths a, b, and c, then the formula for the perimeter is 

p = a + b + c. 

a . Find p when a = 15, b = 40, c = 27. 

b. Find a when p = 67, b = 20, c = 30. 

c. Find b when p = 96, a = 22, c — 43. g 

d. Find b when p = 96, a — 22, c = 39. 

e. If p and a have fixed values, how does b change when c 
decreases ? 

/. If a and b have fixed values, how does c change when p 
increases ? 

2. The formula for the perimeter of a rectangle whose alti¬ 
tude is a and base is b is ^ g 

p = 2a + 2b. 

a. Find p when a = 18 and b = 45. a 

b. Find a when p = 300 and b = 40. _ ? 

c. Find a when p — 275 and b = 40. ^ b ^ 

d. If b has a fixed value, how does a change when p decreases ? 

e. How does p change, if a and b are both doubled ? 

3. Using the formula A = P + PR T : 

a. Find P when A = $3500, R = 5%, and T = 8 yr. 

b. Find R when A = $5000, P = $4000, and T = 2£ yr. 

c. Find T when A = $2600, P = $2000, and R = 6%. 

d. If R and T have fixed values, how does A change when P 
increases ? 

e. If P and T have fixed values, how does A change when 
R increases ? 

4. By the formula S = ^ na + -J- nl: 

a. Find S when n = 10, a = 13, and / = 63. 

b. Find n when S = — 78, a = 19, and l = — 32. 





60 


ALGEBRA 


REVIEW EXERCISE III 

1 . What is the value of — 2 x 2 y when x — 2, y —— 3? 

2 . a. If V = hB, find V when h = 15 and B — 21. 

6 . Find V also when h — 30 and B — 21. 

c. How does F change, if B has a fixed value and h is doubled ? 

d. If B has a fixed value, how does V change if h is halved ? 
If h is trebled ? 

3. Solve the equation f x = 18. 

4. Which of the following is true and which is false? 

a. 3a - 5 + 4 a + 8 — 6 a = a + 3. 

b. — 5x-\-7 + 3x—9 — 2x = 10 x — 2. 

5. a. If l, w, and h represent the length, width, and height of 
a rectangular parallelepiped, and t represents 

the sum of the lengths of all the edges, write 
a formula which expresses t in terms of l, w, b 

and In. 

b. By this formula, find t, when l = 15, 
w = 8, and h = 7. 

c. By this formula, find h, when t = 156, l = 18, and w = 11. 

6 . a. In the figure for Example 5, what is the area of the 
front face? of the back face? of the right-hand face? of the 
left-hand face ? of the lower base ? of the upper base ? 

b. If T represents the total area of all the faces, write a 
formula which expresses T in terms of l, w, and h. 

c. Find T when l — 15, w = 10, and h = 5. 

d. Find l when T = 1160, w = 15, and h = 8. 

e. Find h when T = 695.5, l = 15.5, and w = 8.5. 

7. Twice a certain number is as much more than 7 as five 
times the number is more than 40. Find the number. 

8 . Five sixths of the cost of a certain article is $1.75. What 
is the cost of the article ? 










IV. PARENTHESES 


51. Parentheses are used to group together terms which are 
to be treated as a single number expression. 

Thus, 3 *a — (2 x + y — z) 

means that 2 x + y — z is to be subtracted from 3 a. 

Brackets, [ ], braces, { }, and the vinculum, , are 

used in the same manner. For convenience, they are referred 
to collectively as parentheses, or as symbols of grouping. 

52. Removing parentheses. 

Example 1. x 2 + ( — x + 5) 

means that — x + 5 is to be added to x 2 . 

z 2 .*.** + (-*+ 5) 

— x + 5 = x 2 — x + 5. 

Adding, x 2 — x + 5 

Note. — The parentheses have been removed. The sign of x in the 
result is —, just as it was inside the parentheses; also the sign of 5 
is unchanged. 

Rule. — Parentheses preceded by a plus sign may be re¬ 
moved without changing the signs of the terms which are 
within the parentheses. 

Example 2. x 2 — (— x + 5) 

means that — x + 5 is to be subtracted from x 2 . 

x 2 /. x 2 -(-x + 5) 

- x + 5 = x 2 + x - 5. 

Subtracting, x 2 + x — 5 

Note. — The parentheses have been removed. The sign of x in the 
result is +, whereas it was — inside the parentheses; the sign of 5 in 
the result is —, whereas it was + inside the parentheses. 

Rule. — Parentheses preceded by a minus sign may be re¬ 
moved provided the signs of all terms which are within the 

parentheses are changed from + to —, or from — to +. 

61 





62 


ALGEBRA 


Example 3. Remove the parentheses and collect like terms 
in 2 a — 6 b + (4 a — b) — (5 a — 4 b). 

Solution. 1.4 a and — b, inclosed by parentheses preceded by a 
plus sign, do not have their signs changed when the parentheses are 
removed. 

5 a, which means +5 a, and — 4 b, inclosed in parentheses preceded 
by a minus sign, must have their signs changed. 

2. .*. 2 a — 6 b + (4 a — b) — (5 a — 4 b) 

= 2 a — 6 6 -f 4 a — b — £ a + 4 5. 

3. Collecting like terms, the result is a — 3 b. 

EXERCISE 35 

Remove symbols of grouping and collect like terms: 


1. 

3 m — (m + 4) 

6. 

7a 

+ 

(— 2 a + 6 ) 

2. 

4 x + (2 

— 3 x) 

7. 

6 r 

- 

[- s + 3r] 

3. 

62 / — (5 

- 4-y) 

8. 

8 c 

+ 

£ — 3 c + 2 b\ 

4. 

It - (- 

Zt -2) 

9. 

9 k 

- 

[3 k - 1] 

5. 

8ra + [• 

- 2 m — 5} 

10. 

10 

m ■ 

- 5 — 3 n + 5 m 


Ex. No. 

Remove Parentheses 
from 

Find the Value of 
the Result 

11. 

3 m + (5 m — 2) 

when m = 2 

12. 

7 x — (4 x + 3) 

when x = 5 

13. 

5 t — (6 — t) 

when t =2 

14. 

2 2/ + (- 7 + y) 

when y = 3 

16. 

Z z — ( — 6 + z) 

when 2=4 


16. 

(4 c 

+ 3 d) - (2 c - d) 

21. 

\2a ■ 

— 3 6| 

— {5 a — 

6i>j 

17. 

(x - 

-22/) 

- (x + 3 y) 

22. 

[-a 

+ b]- 

-[c-d\ 


18. 

(2 r 

+ s) 

— (r — 4 s) 

23. 

(-a 

-b)- 

f (-C- 

d) 

19. 

M- 

•«] - 

[t + v] 

24. 

(a 2 - 

■ a ) - 

(a 2 + a) 


20. 

(x - 

- y) - 

1 

?55 

1 

25. 

(5 x 

-8) - 

- (3 x + ■ 

4) 


Historical Note. — Parentheses were introduced by Girard, a 
Dutch mathematician, about 1629. Previously, the brackets and the 
braces had been used by Vieta, about 1593, and by Bombelli, an Italian, 
about 1572. Descartes used the vinculum. 







PARENTHESES 


63 


53. Removing parentheses which are within other paren¬ 
theses. 

Example. Simplify 6 x — (3 x + {2 x — 4} — Jo; — 5 j). 

Solution. 1. The braces inclosing 2 x — 4 and x — 5 are removed 
first, giving attention only to the sign + before (2 x — 4} and the 
sign — before {x — 5}. The rest of the expression is merely rewritten. 

6 x -(3 x + (2 x - 4} - {* - 5}) 

2. Removing {} = 6z— (3z + 2:r — 4 — x + 5) 

3. C. T. inside () = 6x — (4x + 1) 

4. Removing () — 6x — 4x — 1, or 2a; — 1. 

Rule. — To simplify an expression having parentheses within 

parentheses : 

1. Remove the innermost parentheses first, according to the 
rules in § 52. 

2. Combine the terms within the resulting innermost paren¬ 
theses. Then remove these parentheses. 

3. Continue until all parentheses are removed and like terms 
are combined. 


EXERCISE 36 

Remove the parentheses 
1. x + [x +(5 — #)] 


2- y + [y ~(y + 2)] 

3. z + [z — (a — 3)] 

4. w — [x +(3 — x)] 

6. m — [y — {y — 4)] 

6. t — [a — (a + 1)] 

7. ' 2 x + [x — (1 + x)] 

8. 3 y- [y — (3 — 2 2 /)] 

9. 8 s + [5 r +(r — 7 s)] 
10. 3 x — [7 x — |4 x + If] 


11. 9y - (2y - 4-7 y}) 

12. 5r+ [- 4 —(4 r- 3)] 

13. 2 — j6x + (1 - 5x)| 

14. 9 < — |4 +(7< — 3)| 

15. 4 m — [3 m — (m + 1)] 

16. 6 w — 14 — (5 w + 4)f 

17. 5 A — [2 A -(1 - 3 A)] 

18. 8x+|5x — (2 + 5x)j 

19. 2+J3y — (4jf — 3)1 

20. 15 — J8 — (y — 7)J 


21. (7 t- 5) —(3 t —\2t + 1|) 

22. (3 m — 4) — (— 2 m + [5 m — 6]) 

Note. —Exercises 158-159, page 281, can be done now. 


64 


ALGEBRA 


64. Placing terms inside parentheses. 

Development. 1. What is the rule for removing paren¬ 
theses preceded by a plus sign ? 

2. What, then, should be done with the signs of terms which 
are 'placed within parentheses preceded by a plus sign ? 

Example 1. Place the last two terms ofa + 6—c + din 
parentheses preceded by a plus sign. 

Solution. a b — c + d = a + 6 + (— c+d). 

Check. Gt —& —|— (— c d) = a b — c -J- d. § 52 

3. What is the rule for removing parentheses preceded by a 
minus sign? 

4. What, then, should be done with the signs of terms which 
are placed within parentheses preceded by a minus sign ? 

Example 2. Inclose the last two terms of a — b — c -f d 
in parentheses preceded by a minus sign. 

Solution. The signs of — c and + d must be changed. 

••• a-b-c+d = a- b-{+c-d). 

Check. a — b — (c — d) = a — b — c + d. §52 

Rule. — 1. If terms are placed within parentheses preceded 
by a plus sign, their signs must not be changed. 

2. If terms are placed within parentheses preceded by a 
minus sign, their signs must be changed. 

EXERCISE 37 

Inclose the last three terms of the following expressions in 
parentheses preceded: a. by a plus sign; b. by a minus sign. 


1. 

x y + z — w 

8. 

4 m 2 

— n 2 — 6 n — 9 

2. 

r + s — t + v 

9. 

x 2 - 

4 ab — a 2 — 4 b 2 

3. 

m — n + p — g 

10. 

m — 

nx — px — q 

4. 

a — b — c — d 

11. 

r — 

sx — tx — px 

6. 

x 2 + x — 5 — y 

12. 

m 2 - 

- n 2 + 2 nx — x 2 

6. 

a 2 — b 2 — 2 be — c 2 

13. 

a 2 - 

b 2 x 2 + c 2 x 2 — 2 bex 2 

7. 

x 2 y 2 — 2 yz z 2 

14. 

A + By - Cy + Dy 


PARENTHESES 


65 


55. Using parentheses in equations. 

Example. Solve and check the equation 

(8 1 - 11) - (7 - 5 t) = 12 - (13 + 4 <)• 

Solution. 1. (8 i - 11) - (7 - 5 f) = 12 - (13 + 4«). 

2. Removing () 8 £ — 11— 7 + 50=12 — 13— it. 

3. C. T. 13 f - 18 = - 1 - 4 f. 

4. A 18 13 t = 17 - it. 

5. Au 17 1= 17, or t = 1. 

Check.. Does (8 - 11) - (7 - 5) = 12 - (13 + 4)? 

Does - 3 - 2 = 12 - 17? Does - 5 = - 5? Yes. 

EXERCISE 38 

Solve and check the following equations: 

1. 3 x - (x - 4) = 14 4. 9 z - (z - 24) = 72 

2. 4 y - (3 y + 8) = 3 5. 8 c - (11 + 6 c) = - 2 

3. 2 r - (r + 6) = 11 6. (3 y - 4) - (7 + y) = 7 

7. (3 z + 4) - (2z + 9) = 15 

8. (5 t — 8) — (7 — 10 1) = — 12 

9. (2 W - 13) = 18 - (W + 4) 

10. 38 - (9 r + 5) = (6 r - 17) 

11. 6 z + (7 z — 11) = 15 z — (4 z + 5) 

12. (4 m — 13) — (5 + 2 m) = (1 — 4 m) - (30 — 7 m) 

13. (10 t + 3) — '(8 1 - 7) = (15 - 6 t) - (14 t - 6) 

14. (3 A — 4) — (2 A — 10) = (15 — 8 A) — (1 — 5 A) 

15. (6 W - 5) - (9 W - 4) = (6 - 12 W) - (15 W - 1) 

16. (17 - 2 s) - (3 s - 15) + (4 s - 13) = 3 s - 21 

17. (5 r - 4) - (11 - 15 r) = (25 r - 3) - (10 r + 7) 

18. (4 c + 7) - (5 - 8 c) = 20 c - (12 c - 5) 

19. 12 A + (2 A — 3) = 6 A — (— 7 A — 15) 

20. (4 t — 5) — (2 t + 5) = (2 t - 9) — (4 t — 7) 

21. (- 10z + 12) — (6 a; - 5) = (13 - 8 s) - (15 +9x) 

22. (3 x + 5) - (2 x - 7) = (4 x + 9) - (2 x - 11) 

23. (6 y — 8) — (9 y + 4) = (9 — 13 y) — 11 y 


66 


ALGEBRA 


56. Using parentheses to represent number relations. 

Example. Indicate and then simplify the excess of 7 £ over 
the binomial (3 z — 4). 

Solution. 1. “Excess over” means “amount more than.” 

.•. the excess of 12 over 9 is 12 — 9 or 3. 

2. Thus, the excess of 7 x over (3 x — 4) = 7 x — (3 x — 4). 

3. Removing ( ) the excess = 7 x — 3x + 4, or 4 z + 4. 

EXERCISE 39 

Indicate and then simplify, if possible: 

1. The sum of x and (x + 7); of £ and (x — 5). 

2. (2 x + 5) decreased by (4 — x). 

3. (3 m— 8) decreased by (2 m + 5). 

4. The remainder when (2 x — 3) is subtracted from 
(5 x + 6). 

5. The result when (3 + y) is diminished by ( y — 9). 

6. The amount by which (5 r — 4) exceeds (12 r + 3). 

7. The excess of 18 over x ; of 18 over {x — 5). 

8. The excess of (3 z + 4) over (2 z — 8). 

9. The smaller part of 18 if the larger part is l. 

10. The larger part of 20 if (x + 3) is the smaller part. 

11. The difference of 3 a and (a + 7). 

Hint. — Let “difference of” mean the first number minus the second. 

12. The difference of (3 a + 5) and (2 a — 6). 

13. The amount by which 15 is greater than 12; greater 
than a; greater than (a — 2). 

14. The amount by which 20 exceeds 15; exceeds (i — 3). 

15. The amount by which (2 a + 3) exceeds (a — 5). 

16. The smaller part of x if 7 is the larger part. 

17. The larger part of 65 if x is the smaller part. 

18. The excess of (5 m — 8) over (2 m -J- 3). 

19. The sum of two numbers is 50. The smaller is s. Repre¬ 
sent the larger. 


PARENTHESES 


67 


20. a. 5, 6, and 7 are three consecutive integers. What are 
the two larger consecutive integers ? 

b. What must you do to an integer to get the larger con¬ 
secutive integer? 

c. If x is an integer, what is the first larger consecutive 
integer ? the second ? the third ? 

21. a. 2, 4, and 6 are consecutive even integers. What are 
the two larger even integers consecutive to them ? 

b. If z represents an even integer, what are the two larger 
consecutive even integers ? 

22. a. 7, 9, 11, 13 are four consecutive odd integers. What 
are the two larger odd integers consecutive to them ? 

b . If x is an odd integer, what are the two larger consecutive 
odd integers ? 

c. What are these three integers if x is 21 ? 

d. What are these three integers if x is — 5 ? 

23. a. Let x = an integer. Then represent the two larger 
consecutive integers. 

b. Represent the sum of the three integers obtained in part a. 

24. a. Represent three consecutive even integers. 

b. Represent three consecutive odd integers. 

25. Let a equal the altitude of a certain rectangle. 

a. If the base is 4 in. more than the altitude, represent the base. 

b. Represent the perimeter of this rectangle. 

c. How much is this perimeter when a = 5? 

26. The shortest side of a certain triangle is x in. long. The 
second side is 8 in. longer than the first, and the third side is 
3 in. less than twice the first. 

a. Represent the second and third sides and the perimeter of 
this triangle. 

b. What is this perimeter if x = 12 ? 

27. One number is 650 less than a second number. Repre¬ 
sent each of the numbers, and also the sum of the two numbers. 


68 


ALGEBRA 


57. Using parentheses to solve problems. 

Example. Separate 140 into three numbers such that the 
second is 15 less than the first and the third exceeds twice the 
first by 35. 

Solution. 1. Three numbers are to be found. Their sum is 140. 

2. Let / = the first number. 

3. .*. / — 15 = the second, since it is 15 less than the first. 

4. And 2 / + 35 = the third, since it exceeds twice the first by 35. 

5. .*./+(/- 15) + (2/ + 35) = 140. 

6. Removing () / + / - 15 + 2/ + 35 = 140. 

7. C. T. 4/ + 20 = 140. 


4/= 120. 

/ = 30, the first number. 


8 . S20 

9. D 4 
10 . 

11 . 


’. / — 15 = 15, the second number. 
2 / + 35 = 95, the third number. 


Check. 1. The second is 15 less than the first, and the third exceeds 
twice the first by 35. 

2. Does 30 + 15 + 95 = 140? Yes. .". 30, 15, and 95 are the re¬ 
quired numbers. 

Note. — Use parentheses as in Step 5. Read Step 5 thus: 

/ plus the binomial / — 15 plus the binomial 2/ + 35 equals 140. 


EXERCISE 40 


1. One number is 7 more than a smaller number. The sum 
of the numbers is 35. What are the numbers ? 

2. One number exceeds another number by 13. Their sum 
is 25. What are the numbers ? 

3. Separate 100 into two parts such that the larger is 14 
more than the smaller. What are the numbers ? 

4. Separate 124 into two parts such that the greater exceeds 
the smaller by 56. 

5. A boy has a board 36 in. long. He wants to cut it into 
two pieces such that one will be 8 in. longer than the other. 
How long will the pieces be ? 

6. There are two consecutive integers whose sum is 49. 
What are they? (Review Examples 20 and 23, page 67.) 


PARENTHESES 


69 


7. There are three consecutive integers whose sum is 126. 
What are they ? 

8. The base of a certain rectangle is 5 in. longer than the 
altitude. The perimeter of the rectangle is 66 in. Find the 
base and altitude. 

9. The perimeter of a certain triangle is 44 in. The second 
side is 1 in. less than twice the first side; the third side is 5 in. 
more than twice the first side. How long is each side ? 

10. The sum of the ages of three children of a certain family 
is 30 years. Charles’ age is 1 yr. more than twice John’s age; 
Mary’s age is 1 yr. less than 3 times John’s age. Find their 
ages. 

11. The total height of the Library of Congress at Washington 
is 64 ft. The height of its second story is 7 ft. less than twice 
the height of its first story; the height of its third story is 1 ft. 
more than twice the height of its first story. What is the height 
of each of its three stories ? 

12. There are two consecutive odd integers whose sum is 144. 
What are they? (Review Example 22, page 67.) 

13. Separate 75.6 into two parts such that the larger is 49 
more than the smaller. 

14. Separate 55.5 into two parts such that the larger exceeds 
the smaller by 17.9. 

15. A board 14 ft. long is to be cut into two pieces such that 
one will be lj ft. longer than the other. How long will each 
piece be? 

16. The base of a certain rectangle is 4 in. less than 5 times 
the altitude. The perimeter of the rectangle is 1 in. more than 
9 times the altitude. What are the base and altitude ? 

17. The sum of certain three consecutive integers is 204. 
What are they? 

18. Repeat Example 17, letting the integers be even integers. 


70 


ALGEBRA 


19. If the smallest of three consecutive even integers is in¬ 
creased by the second, the sum is 20 more than the third integer. 
What are the integers ? 

20. Divide $175 among three children so that the first will 
receive $10 less than the second, and the third will receive $20 
more than the second. 



21. Charles’ age is 2 yr. more than twice John’s age. Henry’s 
age is 1 yr. more than three times John’s age. The sum of their 
ages is 33 years. How old is each ? 

22. If the angles of a triangle are 
torn off and placed side by side, their 
sum is found to be 180°. 

Memorize this fact. 

How large is each angle of a triangle 
if the second is equal to the first, and 
the third is 3 times the first? 

23. In a triangle ABC, angle B is 20° more than angle A, 
and angle C is 5° less than angle A. How large is each? 

24. If angle A of triangle ABC is 19° more than angle B, 
and angle C is 16° less than angle B, how large is each angle? 




25. There are five consecutive integers such that the sum of 
the first, third, and fifth exceeds the sum of the second and 
fourth by 25. What are the integers ? 


26. There are three numbers whose sum is 360. The second 
exceeds 3 times the first by 45, and the third is 15 less than twice 
the first. What are the numbers ? 


27. Find three numbers whose sum is 423.4. The second is 
18.5 more than the third, and the first is 26.5 less than the third. 

28. The total population of Albany, Buffalo, and Rochester 
was 917,000 in 1920. Rochester’s population exceeded twice 
that of Albany by 69,000; Buffalo’s exceeded four times that 
of Albany by 54,000. What was the population of each ? 




PARENTHESES 


71 


REVIEW EXERCISE IV 

1 . By the formula V = \b + c + 4 m) find V when h = 15, 

6 

b = 16, c = 10, and m = 13. 

2. By the formula S = % gt 2 , find S when g = 32.16 and 
t = 2 . Also when t = 0,1, 3 , 4, and 5 . 

3. Does (- 1)7 = - 7 ? 

4. a. What is the sum of 5 x 2 — 3 x + 7 and 2 x 2 + 4 £ — 9? 

b. What is the value of the sum when x = 3 ? 

5. Subtract x 2 — ^ xy + -J- y 2 from ^ x 2 — xy + § y 2 . 

6 . Simplify 8 a — [ 3 a — J5 a — 3j]. 

7. Subtract 2m— 5 n -j- 8 p from 5m— n -f- 2 p. 

8. From the sum of 8 X 2 — 4 XY — 7 Y 2 and 5 X 2 + 3 XY 
— 2 Y 2 subtract — 2 X 2 — 3 X Y + 5 F 2 . Check, letting 
X = 2, Y = 3. 

9. Solve the equation 8 x — [(7 — 5 x)— 12.04] = 44.04. 

10 . Eight times a certain number exceeds 56 by an amount 
which equals six times the number. What is the number ? 

11. Inclose the last three terms of 16 x 2 — a 2 + 2 ab — b 2 in 
parentheses preceded by a minus sign. 

12 . a. What is the cost in dollars of n dozen of eggs at m 
cents per dozen ? 

b. How does the cost change if n has a fixed value and m is 
doubled ? 

13. a. If a man earns E dollars a week and spends S dollars, 
saving all the rest, how much will he save in one year ? 

b. If E has a fixed value, and S increases, how does the 
amount he saves change ? 

14. If S = C + G, does C = $15.90 when S = $13.65 and 
G = $2.25 ? 


V. MULTIPLICATION 


58. The law of exponents for multiplication. 

Development. 1 . Review the definitions of exponent , base , 
and power of a number, given in § 25 b, page 30. 

2. What does x 2 mean? a 3 ? r 7 ? 

3. Write in exponent form : 

(a) b • b • b • b (6) m • to • to • to • m (c) r • r • r • r • r 

(d) x • x • x ... x (if there are 11 factors) 

4. The product of a 3 and a 4 is found as follows: 

a? = a • a • a; a 4 = a • a • a • a. 
a 3 • a 4 = (a • a • a) X (a • a • a • a) = a • a • a • a • a • a • a = a 7 . 

5. Find, as in Step 4, the following products : 

(a) x 2 • x 4 = ? (6) r 3 • r 5 = ? (c) m 4 • m 5 = ? 

6. Observing the results in Step 5, get the following products 
mentally; check as in Step 4: 

(a) m 2 • m 3 = ? (b) a z • a 6 = ? (c) x 2 • x 5 = ? 

Rule. — To find the exponent of any number in a product, 
add the exponents of that number in the multiplicand and 
multiplier. 

EXERCISE 41 

Using the above rule, give the following products: 


1. 

X 4 

• X 3 

6. 

k 8 • 

¥ 

11. 

w A - 

■ w 8 

16. 

x 9 • 

X 6 

2. 

2/ 6 

• y h 

7. 

m 10 

• m 3 

12. 

c 10 - 

• c 5 

17. 

y A • 

y 20 

3. 

Z 3 -' 

•z 7 

8. 

s 9 - 

s 5 

13. 

f 12 , 

, r 6 

18. 

z m 

•z 2 

4. 

f 4 • 


9. 

w 8 ■ 

• w 7 

14. 

A 9 


19. 

x a • 

X 5 

5. 

z 5 

•z 6 

10. 

a 9 • 

a 11 

15. 

J5 8 • 

B 7 

20. 

x m 

• x n 


72 


MULTIPLICATION 


73 


59. Multiplication of monomials. 

Example 1 . (— 5 x) • (— 6 x 2 y)=(— 5) • x • (— 6 ) • x 2 y 

= (- 5)(- 6) • x • x 2 y 
= + 30 x*y. 

Rule. — To find the product of two monomials: 

1. Find the product of the numerical coefficients, using the 
law of signs for multiplication. (§ 32, p. 38.) 

2. Multiply this product by the literal factors, giving to each 
an exponent equal to its exponent in the multiplicand plus its 
exponent in the multiplier. 

Example 2. Multiply — 5 x 3 yz 5 by +'6 xy 2 w. 

Solution. (— 5 x 3 yz 5 ) (+ 6 xy 2 w) = — 30 x^ +v )y^ +2 h 6 w 
= — 30 x^y^w. 

EXERCISE 42 

Find the following products : 


1. 

a 3 ■ 

x 4 

14. 

(- 

6 a 3 6 4 )(— a 3 b) 

2. 

(- 

2 x) ’ (+ 3 x) 

15. 

(+ 

8 ax 2 ){— 12 ax) 

3. 

(- 

4 y)(— 5 y) 

16. 

(+ 

9 a+/ 4 )(— 9 xy 2 ) 

4. 

(+ 

z){-Zz 2 ) 

17. 

(- 

7 AB 2 ){+ 13 A 2 B) 

6 . 

(- 

w)(— 8 w 2 ) 

18. 

(- 

14 m 2 n){— n ) 

6. 

(- 

xy)(+ xz) 

19. 

(- 

3 a)(+ | a 2 ) 

7. 

(- 

3 oi)(+ 2 a 2 ) 

20. 

(- 

12 )(+f r) 

8. 

(- 

4 xy)\— 6 x 2 y) 

21. 

(- 

15xi/)(- fx 2 ) 

9. 

(- 

8 ab) (+ 3 a 2 ) 

22. 

(- 

40 )(+ft) 

10. 

(- 

3 r 2 s)(— 4 rs 2 ) 

23. 

(- 

i ab) ' (i a 2 ) 

11. 

(+ 

5 xy 2 )(- 8 x 2 ) 

24. 

(- 

-| m 2 ) 

12. 

(- 

6 c 2 d 3 )(— 4 cd ) 

25. 

(- 

f x 2 )(+ | -xy) 

13. 

(- 

9 m 2 n 3 ) (+12 mn 2 ) 

26. 

(.5 

to) • (— 3 TO 2 ) 

27. 

Find: (x 3 ) 2 ; (-5x) 2 ; 

(+ 2 xyY ; 

(- 3 x 2 ) 2 . 

28. 

Find: (x 2 ) 3 ; (-2i) s ; 

(- 3 a 2 

:)8. 

(- i m) 3 . 

29. 

If the base of a rectangle is 5 £ 

in., 

and its altitude is 3 


in., what is its area? 


74 


ALGEBRA 


60. Multiplying polynomials by monomials. 

Observe that 5(7 + 3)= 5 X 10 , or 50 ; 

Also that (5 X 7)+(5 X 3) = 35 + 15, or 50. 

We conclude that 5(7 + 3) = (5 X 7)+(5 X 3) 

Similarly a(b + c + d)= ab + ac + ad 

Rule. — To multiply a polynomial by a monomial, multiply 
each term of the polynomial by the monomial and combine the 
results. 

Example. Multiply 3 a 2 — 2 ab + b 2 by — 3 ab. 

Solution. 

(— 3 ab) X (3 a 2 — 2 ab + 6 2 ) = — 9 a 3 b + 6 a 2 6 2 — 3 ab 3 . 

Note. — In this solution, (— 3 ab) (3 a 2 ) = — 9 a 3 6 ; (— 3 ab) (— 2 ab) 
= + 6 a 2 6 2 ; and (— 3 ab) ( b 2 ) = — 3 ab 3 . All this should be done 
mentally. 

Check. If a = 1 and b = 2, —3 ab = — 6 ; 

3 a 2 — 2 ab + 6 2 = 3 — 4 + 4, or 3; and (— 6 ) X (3) = — 18. 
Also, the result, — 9 a 3 b + 6 a 2 6 2 — 3 ab 3 = — 18 + 24 — 24, or — 18. 


EXERCISE 43 


Multiply: 

1. 5 a — 8 by 3 a 

2. x 2 — 4 x + 5 by 2 x 

3. 7 xy{x - 4 y) 

4. 6(| z -1 y) 

6.-4 a(3 z 2 + 2 z — 10) 

6 . 5 a 2 (2 a 3 - 5 a 2 + 7) 

7. -3 s(r 2 -2rs + s 2 ) 

8 . -3z 2 (6z 4 - 7z 3 + 5z 2 ) 

17. 5 m 2 — 4 mn + 3 n 2 
2 m 

18. 7 a 2 — 5 ab + 6 6 2 
— 3 ab 


9. — 2 c 2 ( — 2 c 2 -f 3 cd + 4 d 2 ) 

10. - x z (x h - x 4 + 5 x 2 ) 

11 . — 6 a6(10 a 3 — 5 b 2 ) 

12 . x 2 y (x 3 — 3 x 2 y + £y 2 ) 

13. — rs(r 2 + rs + s 2 ) 

14. — 5 ab(a 2 — 3 ab — 7 b 2 ) 

15. -2 mn(4: m 2 — 3 mw + 2 w 2 ) 

16. 6 a 3 (3 a 2 — 4 + 6 2 ) 

19. 2 a : 2 - 3 xy + 4 y 2 

— 4 xy 

20 . ^ - ^ + 3x - 4 

— 5 x 






MULTIPLICATION 75 


21 . Multiply f t 4* — f t by 60, collect terms, and find 

the value of the result when t = 2 . 

/o -t o \ 12 q 30 1 20 cy 

Solution. 1 . 60(|z + ~t - ^t) = 00 + 00 \t - 00 •?* 

\o z 6 / & 2 3 

2. = 36 t + 30 * - 40 * 

3. = 26 Z. 

4. When Z = 2, 26 £ = 26 • 2, or 52. 

Multiply as in Example 21 , and find the value of: 

22 . 30 (^1 — tV t + i t) when t = 10 . 

23. 40(f x — x — ^ x) when x = — 4. 

24. 36 ( 325 - w ~ 1 w — i) when w = 8. 

25. 28(i^- a + f 6 — ^ c) when a = 2, b = 3, and c = 9. 

26. a. The base of a certain rectangle is (x + 8 ) in., and its 

altitude is 3 £ in. What is its area ? 

b. What is this area when x = 2 ? 

27. The altitude of a certain triangle is (y — 3) in., and its 
base is 12 y in. What is its area ? 

28. There are n articles which cost (n + 15) cents each. 
What is their total value ? 

29. A train travels t hours at the rate of (10 t -f 5) miles 
per hour. How far does it travel ? 

30. Simplify 5 y — 3(x — 4). 

Solution. 1. 3 times (x — 4) is to be subtracted from 5 y. 

2. But, by § 31, this is the same as adding — 3(x — 4) to 5 y. 

3. simply multiply (x — 4) by — 3 and write the results. Then 

5 y - 3{x — 4) = 5 y — 3 z + 12 


Simplify the following expressions, combining like terms. 


31. 5 y - 2(3 y - 4) 

32. 6 s + 8(4 s - 2) 

33. 7 1 - 5(6 - 3 <) 

34. — 8 w + 9(2 w — 4) 
36. - 13 x - 9(5 - 3 x) 


36. (3 z — 7) — 6(4 z — 10) 

37. 9(11 —7 k) -8(12 —5 k) 

38. 12(1 m + 4) + 9(1 m - 5) 

39. 10(1 — 3s)— 8(4 —5 s) 

40. 16(fc + f) — 6(2 c -3) 


76 


ALGEBRA 


61. Multiplying a polynomial by a polynomial. 

The product (;x -f y)(b + c) may be rep- ^ 
resented by the figure at the right, (x + y) 
is the base, and (b + c ) is the altitude, c 
The areas of the four parts are marked. 

(x + y)(b + c) = bx + by + cx + cy 


bx 

by 

cx 

cy 


x y 


Rule. — To multiply one polynomial by another : 

1 . Arrange multiplier and multiplicand in ascending or 
descending powers of the same letter. 

2 . Multiply the multiplicand by each term of the multiplier. 

3. Add the partial products. 

Example . Multiply 3a — 46 by 2a — 5 6 , and check the 

solution by letting a = 1 and b = 2 . 

Solution. 3 a — 4 b 

2a - 5b 

6 a 2 — Sab This is 2 a X (3 a — 4 b) 

— 15 ab + 20 b 2 This is — 5 b X (3 a — 4 b) 

Adding, 6 a 2 — 23 ab + 20 b 2 This is the product. 

Check. a and b may be any numbers. 

When a = 1 and b = 2, 3 a - 4 b = 3 - 8, or - 5. \ And (— 5) (— 8) 

2 a — 5 b = 2 — 10, or — 8. J = + 40. 

The product, 6 a 2 — 23 ab + 20 b 2 = 6 — 46 + 80, or 40. 

Since the two results are equal, the product is probably correct. 


EXERCISE 44 


Multiply: 

1 . x + 9 by x + 7 

2 . z + 11 by z + 13 

3. y — 7 by y — 5 

4. z — 8 by z — 10 
6 . w + 4 by w — 12 

6 . w — 16 by w + 3 

7. 22 + 5 by 32+2 


8 . 6 m—3by4m + 5 

9. 4 t — 5 by 6 t — 2 

10 . 8 a; — 3 by 5 z - 4 

11. 3 a + 2 b by 2 a — b 

12 . 5m — 6fby3m + 4f 

13. 7w + 6 pby 8 w— 4p 

14. 9 z 2 - y 2 by 2 z 2 + 3 y 2 







MULTIPLICATION 


77 


15. 

(3 w 1 — 

■ 7 m) (4 w 2 + 5 m) 

25. 

(2 x - 

■ 3 y)(.5 x + .3 y) 

16. 

(i a + 

ib)(2a + b) 

26. 

(.4a+ .5 6)(.4o+ .5 b ) 

17. 

(i*- 

iy)(ix + iy) 

27. 

(* 2 + 

2*- 3)(*- 1 ) 

18. 

(f x ~~ 

CO 

«|to 

1 

CO 

fca 

28. 

(x 2 + 

x v + y 2 )(x - y) 

19. 

(f m - 


29. 

(x 2 — 

xy + y 2 )(x + y) 

20. 

(ir x “b 

i y)d* — iy) 

30. 

(m 2 — 

4 m + 3)(m — 2 ) 

21. 

(2 x 2 - 

3 y)(x 2 + 7 y 2 ) 

31. 

(a 2 - 

2 a — 4) (2 a -f- 3) 

22. 

(m + n)(r — s) 

32. 

(3 r 2 + 6 r - 8 ) (4 r + 3) 

23. 

( 2 c + 

3 d) (a — 6 ) 

33. 

(5 x 2 - 

-2x — 7) (3 x — 2) 

24. 

(3 a — 

4 6 ) (c + d) 

34. 

(3 + 2 

ly-4y>X2-3y) 

35. 

(4 c 2 - 

cd+ d?)(2c- d) 




36. 

(m 2 — 

3 mn — w 2 ) (m + 2 n) 



37. 

(r 2 - 6 

' rs — 2 s 2 ) (r — 4 5 ) 




38. 

(4 to 2 - 

- 2 mn + n 2 ) (m — 3 

n) 



39. 

(4 s 2 + 

r 2 - 4 rs) (r -2 s) 

Check for 

r = 2, s = 3. 


40. (3 c 2 + d? — 2cd)(2c+ d) Check for c = 2 , d = — 2 . 

41. (4 n 2 — 7 + 5 n)(2 n — 3) Check for n = 2 . 

42. (5 t + 2 £ 2 — 6 ) (3 t — 1 ) Check for t = — 2 . 

43. a. What is the area of the rectangle whose base is 
(5 x + y) in., and whose altitude is (2 x — y) in. ? 

6 . How much is this area when x = 3 and y = 4? 

44. a. What is the volume of the cube whose edge is (2 a — b) 
in. ? b. What is this volume when a = 4 and 6 = 3? 

45. a. What is the cost of (2 n m) articles at (n — m)fc 

each ? 6 . How much is this result when n = 8 and m — 3 ? 

46. a. A man traveled (n + 8 ) hr. at the rate of (m + 2 ) mi. 
per hour. How far did he go ? 

6 . What were his rate, time, and distance when m = 30 and 
n = 3? 

Note. — Exercise 160, p. 282, can be done now. 


78 


ALGEBRA 


62. Multiplication used in solving equations. 

Recall the rules used in solving equations, taught in § 13, 
§ 43, and § 47. These enable you to understand: 

Rule IV. Both members of an equation can be multiplied 
by the same number without destroying the equality. 

Example. Solve the equation 7(90 — a)— 2(180 — a) = 20. 
Solution. 1. 7(90 — a) — 2(180 — a) = 20. 

2. Multiplying, 630 — 7a — 360 + 2 a = 20. 

3. C. T. - 5 a + 270 = 20. 

4. S 270 — 5 a = — 250. 

5. M_x 5 a = 250. 

6. D 5 a = 50. 

Check. Does 7(90 - 50) - 2(180 - 50) = 20? 

Does 7 • 40 - 2 • 130 = 20? Does 280 - 260 = 20? Yes. 

Note. — In Step 2, - 2 • 180 = - 360 ; - 2( - a) = + 2 a. 

In Step 5, what does M_i mean? Recall also, Ex. 30, p. 75. 

In Step 5, observe how — 5 a is changed to +5 a. 


EXERCISE 45 

Solve and check the following equations: 


1. 3 (a: - 4)= 33 

2. 2(3 y — 8)= 20 

3. 5(2 x — 1)= 35 


7. 4(2/ — 3)— 5 = 3 

8. 6(2 + 2)- 40 = 50 

9. 8(3 t - £)+ 9 = 17 


4. 6(8 x — 1)= 4 x + 16 10. 2(3 r - 11)= 5(11 - r ) 

6 . 7(1 - 9 1)= 12 t - 18 11. 4(16 - m) = 3(m - 9) 

6. 3(1 - 12 ra) = 4 m - 7 12. 2(7 z + 1)= 3(7 x - 4) 

13. 3(2 x — 4)— 2(x — 5)= 2(x + 6) 

14. 6(2 - 3 y)+ 30 = 3(4 y - 6) 

16. 14 - 5(3 a - 2)= 3(o - 10) 

16. 10 - 5(r - 4)= 6(V- - r) 

17. 2(9- 3 *)+ 3(4 — »)= 21 

18. 7(y - 8)- 6(2 g - 15)= 4(6 - g ) 

19. 3(5 - «)- 4(2 n — 7)+ 3(7 + n)= 0 

20. 4(3 m - 5)+ 3(6 m - 1)= 3(9 m - 7)- 1 


MULTIPLICATION 


79 


21. The greater of two numbers is 7 more than the smaller. 
Twice the greater number is 6 more than three times the 
smaller. Find the numbers. 

22 . One number exceeds a certain other number by 5. Six 
times the smaller number, diminished by four times the larger 
number, is 6 . What are the numbers ? 

23. The sum of two numbers is 45. If four times the smaller 
be decreased by twice the larger, the result is 12 . What are the 
numbers ? 

24. One number exceeds another number by 3. Six times 
the larger, diminished by seven times the smaller, is 7. What 
are the numbers ? 

25. There are two consecutive integers such that the sum of 
twice the larger and four times the smaller is 110 . What are 
these integers ? 

26. There are three consecutive integers such that three times 
the smallest, plus two times the second, plus the third, gives 100 . 
Find these integers. 

27. There are three consecutive integers such that the 
smallest, plus twice the second, plus three times the third, gives 
100 . What are they ? 

28. The base of a rectangle is 8 in. more than the side of a 
certain square; its altitude is 3 in. less than the side of the 
square. The perimeter of the rectangle is 74 in. What is the 
length of a side of the square, and what are the base and altitude 
of the rectangle ? 

29. In 1920, the population of the state of New York was 
about 10,400,000. The city population exceeded the rural 
population by about 6,800,000. What was the city and the 
rural population ? 

30. Separate 35 into two parts such that twice the larger 
is 10 more than three times the smaller. 


80 


ALGEBRA 


EXERCISE 46 

Age Problems 

1. A is now 18 years old. Express his age : 

a. 6 years ago; b. n years ago; c. x years from now. 

2. B is x years old. Express his age : 

a. 4 years hence (from now); b. 6 years ago; c. y years ago. 

3. A is now x years old. B is 8 years older. Express : 
a. B’s present age; b. the sum of their ages ; 

c. the age of each 10 years ago; 

d. the age of each 5 years hence. 

4. A is now n years old. B is four times as old. Express: 

а. B’s present age ; b. the age of each 4 years hence; 

c. the fact that B’s age as found in part b is 3 times A’s age. 

d. Solve the equation formed in part c. Obtain the age of each. 

б. B is three times as old as A is now. Make a table and 
supply the answers to the questions asked below. 

a. Let a = the number of years in A’s age now. 



Now 

10 Years Ago 

A’s age 

? 

? 

B’s age 

? 

? 


b. Write the equation which expresses the fact that B’s age 
ten years ago was 5 times A’s age then. Solve it, and obtain 
the ages of A and B. 

c. Check the solution by finding their ages ten years ago, and 
comparing them with the statement in the problem. 

Solve the following problems : 

6 . A is now four times as old as B. In six years he will be 
twice as old as B is then. How old is each now ? 

7. C is now five times as old as D. In fifteen years he will 
be twice as old as D is then. How old is each ? 








MULTIPLICATION 


81 


8. A is now 12 years older than B. Four years from now, 
A will be twice as old as B is then. What are their present ages ? 

9. The sum of the ages of a father and son now is 30 years. 
In 9 years the father’s age will be only 3 times the ston’s age then. 
How old is each now ? 

10. A’s age is now 38 years and B’s is 5 years. 

a. How old will each of them be in x years ? 

b. If A will be 4 times as old as B at that time, form the 
equation from the ages obtained in part a, and solve it for x. 

11. A’s age is now 33 years, and B’s is 3 years. In how many 
years will A’s age be just 6 times B’s age at the same time ? 

12. A’s age now is 65 years, and B’s is 20 years. How many 
years ago was A 6 times as old as B was ? 

13. A is four times as old as B now. Five years ago, he was 
7 times as old as B was then. Find their ages. 

14. A is 5 times as old as B. Five years ago, he was 4 times 
as old as B will be one year hence. Find their ages. 

15. A is now 6 years older than B. Three years ago, A was 
twice as old as B was then. Find their ages. 

16. A is now 10 years older than B. Five years from now 
he will be 5 times as old as B was five years ago. Find their ages. 

17. A is now 7 times as old as B. A’s age in four years will 
be 5 times what B’s age will be in two years. Find their ages. 

18. A is now 4 years older than B. Eight years ago, A was 
twice as old as B was then. Find their ages. 

19. C is now 4 times as old as B. Ten years from now, he 
will be only twice as old as B is then. Find their ages. 

20. The sum of the ages of John and Jim is 20. Four years 
ago, John was twice as old as Jim. How old is each now ? 

21. The sum of the ages of Mary and Edna is 21. Four years 
from now Mary will be as old as Edna was five years ago. How 
old is each now ? 


82 


ALGEBRA 


EXERCISE 47 

Mixture and Coin Problems 

1. A grocer has some 90^ tea and some 40^ tea. How many 
pounds of each must he take to form a mixture of 100 pounds, 
which he can sell for 60^ per pound ? 

Solution. 1. Let n = the number of pounds of the 90^ tea. 

2. .'. 90 n£ = the value of these n pounds. 

3. Then 100 — n = the number of pounds of the 40^ tea. 

4. .*. 40(100 — n)£ = the value of these (100 — n) pounds. 

5. 90 n + 40(100 — n)l = the total value of the tea. 

6. Also 100 X 60j£ = the total value of the tea. 

7. .*. 90 n + 40(100 - n) = 100 X 60. 

Complete and check your solution. 

Note. — The essential statements in Steps 2-6 can be arranged thus : 


Kind of Tea 

No. of Pounds 

Value in Cents 

90 tea 

n 

90 nji 

40^ tea 

100 - n 

40(100 -n)i 

60^ tea 

100 

100 X 60 


2 . A grocer wishes to make a mixture of 60 pounds of coffee 
to sell at 40^ per pound. He has some 55^ coffee and some 30^ 
coffee. How many pounds of each should he take ? 

3. A seedsman wishes to make a mixture consisting of clover 
seed and timothy seed which he can sell at 11^ per pound. His 
clover seed is worth 23^ per pound and his timothy 9^ per pound. 
How many pounds of each should he take for a batch of 200 
pounds ? 

4. A grocer has 20 pounds of 60^ coffee. How many pounds 
of 30^ coffee should he mix with it so that the mixture can be 
sold at 35^ per pound ? 

5. A dealer has walnuts valued at 43^ per pound and almonds 
valued at 35$£ per pound. How many pounds of each should he 
take to make a mixture of 150 pounds to sell at 40^ per pound? 






MULTIPLICATION 


83 


6. A sum of money amounting to $2.80 consists of pennies, 
nickels, and quarters. The number of nickels is 3 times the 
number of pennies, and the number of quarters is 3 more than 
the number of pennies. How many coins of each kind are there ? 

Solution. 1. Let p = the number of pennies, 


Kind op Coin 

Number op Them 

Value op Them 
in Cents 

Pennies 

V 

V 

Nickels 

3 P 

15 p 

Quarters 

P +3 

25(p + 3) 

All together 


280 


3. p + 15 p + 25 (p + 3) = 280. 

Complete the solution, obtaining p. Then answer the question 
asked, and check your solution. 

7. A sum of money amounting to $2.55 consists of nickels 
and dimes. The number of dimes is 6 more than the number 
of nickels. How many coins of each kind are there? 

8. A sum of money amounting to $5.35 consists of nickels, 
dimes, and quarters. There are 5 more quarters than there 
are nickels; the number of dimes is 1 less than three times the 
number of nickels. How many coins of each kind are there ? 

9. A sum of money amounting to $38.75 consists of quarters, 
half-dollars, $1.00 bills, and $2.00 bills. The number of half- 
dollars is 2 more than the number of quarters; the number of 
$1.00 bills is 3 less than twice the number of quarters; the num¬ 
ber of $2.00 bills is 1 less than the number of quarters. How 
many coins of each kind are there ? 

10. A girPs savings bank had in it pennies, nickels, dimes, 
and quarters. Their total value was $10.16. The number of 
dimes was twice the number of quarters. There were 15 more 
nickels than there were dimes. The number of pennies was 3 
less than four times the number of quarters. How many coins 
of each kind were there ? 










84 


ALGEBRA 


EXERCISE 48 

Mixture Problems 

Illustrative Example. How much distilled water must be 
added to 1 qt. of 95% alcohol to reduce it to 40% alcohol ? 

Solution. 1. “95% alcohol” means that, in one quart, there is 
.95 qt. of pure alcohol. 

2. Let w = the number of quarts of water to be added to the 1 qt. 
of 95% alcohol. 

3. (w + 1) = the no. of qt. in the mixture. 

4. A0(w + 1) qt. = pure alcohol, according to the conditions. 

But .95 qt. = pure alcohol, according to step 1. 

5. .400 + 1) = .95 

6. Solving this equation, w = If. 

Hence, If qt. of distilled water must be added. 

1. How much water must be added to one quart of 95% 
alcohol to make a mixture which shall be 50% alcohol ? 

2. Tincture of arnica of standard strength contains 20% 
arnica and 80% alcohol. How much water must be added to 
one quart of 20% arnica to make a mixture containing 15% 
arnica ? 

3. How much alcohol must be added to a pint of 10% pure 
belladonna to make a mixture of 8% pure ? 

4. How much distilled water must be added to two quarts 
of 25% pure rose water to make 10% pure rose water ? 

5. How much distilled water must be added to two quarts 
of chemically pure (100%) hydrochloric acid to make a mixture 
containing 10% of hydrochloric acid ? 

6 . There is a 40 lb. alloy of tin and copper of which 6 lb. are 
tin. How many pounds of copper must be added to make the 
new alloy 5% tin ? 

7. How many quarts of a 4% salt mixture must be added 
to 24 quarts of a 12% salt mixture to give a mixture containing 
10% salt? 


MULTIPLICATION 


85 


REVIEW EXERCISE V 

1. By the formula T = l ~, find T when l = 19, w = 10.5, 
and h = 8. 

2. From the sum of 6 xy — 4 y 2 + 5 x 2 and 2 x 2 — 3 xy + 
5 y 2 , subtract the sum of 3 y 2 - 2 xy + z 2 and 5 z 2 + 7 zy 
- 6 y 2 . 

3. How much less than 2 a 2 — 3 ab + b 2 is a 2 + ab + b 2 ? 
Check by letting, a = 2 and 6=1. 

4. By the formula in Example 1, find h when T = 20, 
l = 10, and w = 6. 

6. Simplify 6 r — 2(— 4 + 5 r)+ 3(— 8 r — 9). 

6. Simplify 5 x — 3(z — 2)+ 2(5 — x). 

7. Solve the equation 17 m — 5(2.4 m — .6)= 18. 

8. Is 5 a root of the equation 11 x — 2(5 x — 1)= 6? 

9. Solve the equation of Example 8. 

10. Multiply 2z 2 — 3 x + lbya: — 3. Check for x = 2. 

11. Simplify 5(3 x — 8 y)— 3(x — 2 y)— 7(x — 5 y). 

12. a. How many cents are there in x nickels, y dimes, and 
z quarters ? 

6. Let C = the result of part a, thus making a formula. 

c. By the formula derived in part 6, find C when x = 3, 
y = 4, and 2 = 6. 

13. The sum of two numbers is 25. Four times the smaller 
exceeds twice the larger by 22. Find the two numbers. 

14. Simplify 3 x — [2 x — (5 — a;)]. 

16. Give the rule expressed by each of the following formulas : 

a. A = hb c. C = 2 irr e. V — Iwh 

b. A = %hb d. I = PRT f. A = Trr 2 


VI. DIVISION 


63. Division is the process of finding one of two numbers 
when their product and the other number are given. 

To divide 15 by 3 means to find the number by which 3 must be 
multiplied to give the product 15. 

The dividend is the product of the numbers ; it is the number 
divided. The divisor is the other given number; it is the 
number by which the dividend is divided. The quotient is the 
required number. 


64. Division is indicated either by the symbol -f- or by 
writing a fraction whose numerator is the dividend and whose 
denominator is the divisor. 

Thus, to divide 15 by 5 is indicated either by 15 -f- 5 or by 

65. If any number (except zero) be divided by itself, the 
quotient is 1 . Thus, 

a -s- a = 1, since a — 1 • a 


66. Division of a product by a number. 

Example. Divide 6 X 8 by 2. 

3 

Solution. 1. 6 may be divided by 2. Then ^ ^ ^ or 24. 

1 2 


4 

2. 8 may be divided by 2. Then 6 (5 8 = 6 * ^ , or 24. 

z z 

3. But both 6 and 8 may not be divided by 2, 


for then 


6X8 

2 


3 4 
6X8 

2 


, or 12. 


Rule. — To divide the product of two or more numbers by a 
number, divide any one of the factors by the number, but divide 
only one of them by it. 


86 








DIVISION 


87 


EXERCISE 49 

Find each of the following indicated divisions in two ways: 

9 X 12 18 X 24 28 X 56 

3 2 ' 6 3 ‘ 7 

67. Division by zero is impossible. 

Thus, if we try to find the quotient of 6 -5- 0, and let q equal the 
quotient, we should have the relation 6 = 0 • q. 

But 0 • q = 0 and not 6, so there cannot be any number to use as q. 

68. The rules of signs for division. 

Since (+2) X (4- 3) = + 6, then (+6) -f- (+ 2) = +3. 

Since (—2) X (+ 3) = — 6, then (— 6) -r (— 2) = + 3. 

Since (+2) X ( — 3) = — 6, then (— 6) -s- (+ 2) = — 3. 

Since (—2) X ( — 3) = +6, then (+ 6) -f- (— 2) = — 3. 

Rule. — 1. The quotient of two numbers having like signs is 
positive. 

2. The quotient of two numbers having unlike signs is nega¬ 
tive. 

EXERCISE 50 

1. Divide -f 6, + 18, — 24, — 60, + 48, — 30, + 72 
a. by + 3 b. by — 2 c. by -f- 6 d. by — 4 e. by - 12 


Divide: 


2. 

- 

18 

by + 3 

12. 

- 

100by - 

-20 

22. 

+ 

17 by ■ 

- 1 

3. 

+ 

12 

by - 

- 2 

13. 

- 

81 by + 

3 

23. 

+ 

1 by - 

-3 

4. 

+ 

15 

by + 5 

14. 

- 

90 by - 

10 

24. 

- 

T 1 by 

- 4 

5. 

— 

30 

by - 

- 6 

15. 

- 

84 by + 

7 

25. 

- 

i by 4- 

i 

6. 

— 

45 

by - 

- 9 

16. 

+ 

60 by — 

10 

26. 

- 

9.6 by 

- 3 

7. 

— 

50 

by - 

- 5 

17. 

- 

10 by - 

20 

27. 

+ 

5.8 by 

- 2 

8. 

+ 

18 

by - 

- 2 

18. 

- 

5 by + 4 

28. 

- 

6.4 by 

- 2 

9. 

— 

24 

by + 3 

19. 

- 

15 by - 

15 

29. 

+ 

2.5 by 

- .5 

10. 

— 

36 

by - 

- 6 

20. 

+ 

32 by — 

32 

30. 

- 

3.6 by 

- .9 

11. 

+ 

42 

by - 

- 7 

21. 

- 

18 by - 

1 

31. 

- 

2.8 by 

+ -7 





88 


ALGEBRA 


69. The law of exponents for division. 

Development. 1. Divide a 5 by a 3 . 

1 1 1 

o 7 .. a 6 • a • a _ 2 

Solution. — =-= 1 • a • a, or a 2 . 

a 3 m; *^r 

Therefore a 5 4- a 3 = a 2 . 

Each a in the denominator is divided into one of the a 's in the 
numerator. The quotient in each case is 1, since a 4- a = 1. 

2. Find as in Step 1 the following quotients : 

a. y 6 4- i/ 4 = ? b. m 6 4- m 4 = ? c. f 10 4- f 6 = ? 

3. Examine carefully the exponents in the dividend, the 
divisor, and the quotient in Step 2. In the following problems, 
try to give the results immediately without going through the 
solution as in Step 1. Check by multiplication. 

a. p s 4- p 6 = ? b. a 6 4- a 2 = ? c. 6 7 4- 6 4 = ? 

Rule. — The exponent of any factor in a quotient is equal to 
its exponent in the dividend minus its exponent in the divisor. 

70. If b is not zero, then b 2 4- b 2 = b 2 ~ 2 = b° by the law 
of exponents. Since b 2 4- b 2 must equal 1, we agree that b° = 1. 

The zero power of any number which is not zero, is 1. 

71. Division of monomials by monomials. 

Rule. — To divide a monomial by a monomial: 

1. Make the quotient positive, if the dividend and divisor 
have like signs; make it negative, if they have unlike signs. 

2. Find the quotient of the absolute values of the numerical 
coefficients. 

3. Multiply the quotient of Step 2 by the product of the literal 
factors, giving each its exponent in the dividend minus its ex¬ 
ponent in the divisor. 

4. Any literal factor which has the same exponent in the 
dividend and divisor gives 1 as a factor in the quotient. 



DIVISION 


89 


Example 1. = - 6 a (7-4)&<2-2) c (3-2). 

= — 6 a 3 6°c, or — 6 a 3 c. 

Example 2. (—33 a?bx 2 y 4 ) -h (+ 3 a B x 2 i/) = — 11 aby 3 . 

Check. Does (+ 3 a 5 z 2 ?/) X (— 11 afo/ 3 ) = — 33 a?bx 2 y A ? Yes. 

EXERCISE 51 

1. 2/ 6 by t/ 4 6. a 3 6 2 by ab 9. a 5 ?/ 3 by x 4 ^ 

2. ra 7 by m 2 6. rV by r 2 s 10. ra 3 n 2 by mn 2 

3. r 8 by r 5 7. z 6 ?/ 3 by x 2 tf 11. 12 a 3 by 3 a 2 


4. 

* 9 by t 4 

8. ra 4 ' 

13. 

20 rV by 5 r 2 * 


14. 

28 c 4 ^ 3 by 7 c 2 ^ 


15. 

— 14 ran by — 

7 n 

16. 

+ 18 a 2 6 by — 

3 ab 

17. 

- 12 xy* by - 

6 xy 2 

18. 

— 24 r*t by + 

Sr 2 

19. 

>> 

xi 

"fe 

CO 

1 

■f 17 ran 3 

20. 

+ 40 x b y 6 by - 

"b 

00 

1 

21. 

— 32 a z b 4 c by ■ 

- 4 ab 4 c 

22. 

35 x 5 y 2 z? by - 

7 o?y 2 

23. 

— 44 a h xy z by 

— 4 a 4 y 2 

24. 

75 x 5 y 4 z? by — 

15 xy 3 

37. 

Divide x m y n by x?y b . 

Solution. 

x m y n 

38. 

X 2r -4- X r 


39. 

X s -4- X 1 


40. 

* - ylm 


41. 

4 x m -4- (— 2 x n 

) 

42. 

x a y b -4- x 3 y 4 


43. 

- 12 x 3a 4-(- 

2z°) 


i 6 by ra 4 n 12 . 18 a; 4 by 6 x 2 

25. 42 aW by - 14 ab 2 c 2 

26. — 64 x 8 y 6 by — 8 £ 7 y 6 

27. — 42 a6 3 c 5 by — 7 b 2 c 4 

28. - 60 £ 4 y 5 z 6 by - 12 x 2 yh* 

29. — 60 a 4 b 5 c? by — 20 a 4 6 3 

30. + 84 c 5 d 6 by - 12 c 4 d 

31. — 51 a&V by — 3 ac 2 

32. + 48 r 5 s 6 by - 16 rV 

33. 5 ab by 10 a 

34. 4 mw 2 by 8 m 

35. — 3 r by — 15 r 

36. + 9 cd by — 12 c 

-J_ _ x m-ayn-h' 

44. — -4- (— a 2 b v ) 

45. — 14 r 4x s 6y -4-(-f 2 r^ y ) 

46. x n+2 y m -4- x 2 y 

47. — x n+1 y n -4- xy 

48. + 24a 2m + 3 -4-(- 4a 2m ) 

49. - 32 x 3a y b+2 ^ (- 16 arV +1 ) 



90 


ALGEBRA 


72. Division of polynomials by monomials. 

Development. 1. Since 2 X 9 = 18, then J^ 8 - = 9. 

2. Since 2(x + 3) = 2 x + 6, then ^ x - = x + 3. 

3. Since 3 (a — 5) = 3 a — 15, what does ^ a ^ equal? 

O 

4. What does each of the following equal? Test the result 
by multiplying the divisor and quotient. 

2 x + 2 i 6 r + 4 6m — 9 

2 2 3 

5. Since a(b + c) = ab + clo, 

Then ab+_ac = 6 + c _ 

a 

Rule. — To divide a polynomial by a monomial: 

1. Divide each term of the dividend by the divisor. 

2. Unite the results with their signs. 

Example 1. Divide 12 a; 3 — 6 x 2 + 3 x by — 3 x. 

Solution. 12 a? - 6 a: 2 + 3 x = _ + 2 x - 1. 

- 3 x 

Check. (- 4 a 2 + 2 * - 1) • (- 3 x) = 12 z 3 - 6 x 2 + 3 s. 
Example 2. Divide — 9 a 3 -f 3 a 2 — 12 a 4 by —3a 2 . 
Solution. (- 12 a 4 - 9 a 3 + 3 a 2 ) (- 3 a 2 ) = 4 a 2 + 3 a - 1. 

Check. Multiply the quotient by the divisor; the result should 
equal the dividend. 

EXERCISE 52 

Divide: 

1. 5 r — 10 s by 5 6. 18 m 3 — 12 m by 3 m 

2 . 12 p — 8 q by 4 6. 6 z 4 — 3 z 3 by — 3 x 

3. 15 x 2 — 18 y 2 by 3 7. — 20 r 3 -f 14 r 2 s by — 2 r 2 

4. 9 x 2 — 7 x by x 8. 12 m 2 n — 18 mn 2 by — 6 mn 

9, - 33 a 2 b + 44 ab 2 by - 11 ab 
10 . 28 c z d - 21 cd? by - 7 cd 









DIVISION 


91 


11. - 72 r 6 - 80 r 4 + 64 r 2 - 32 r by - 8 r 

12. 15 x s — 18 x 2 y + 21 xy 1 2 by — 3 x 

13. - 48 x 10 + 96 z 8 - 72 z 6 by - 12 z 5 

14. 12 y b — 8 y 4 + 16 y 3 6 7 8 9 by — 4 y 2 

16. 60 m 9 — 45 m 6 + -30 m 3 by — 15 m 3 

16. 33 a 3 b 3 c 3 — 44 a 2 6 2 c 2 + 55 abc by — 11 abc 

17. 25 m 7 n & — 30 mV + 40 mV by — 5 vin 2 

18. 15 c 4 d 2 — 20 c 3 d 3 — 25 c 2 d 4 by — 5 c 2 d 2 

19. 6 a 2 b 2 c 3 — 9 a 3 6 3 c 4 +12 a A bc 2 by — 3 cPbc 2 

20. — 49 x 3 y A z 2 + 56 x 2 yV by — 7 x 2 y 2 z 2 


a b b — a z b 3 + ab b 

24. 

— 2 m 2 + 4 mw — Qn 2 

ab 

- 2 

x l - 12 x 3 + 6 x 2 

26. 

16 £ 3 y — 12 a+/ 2 + 8 xy 3 

z 2 

— 4 y 

— m 2 + m — 1 

26. 

— 5 m 3 + 10 m 4 — 15 m 5 

- 1 

— 5 m 3 


REVIEW EXERCISE VI 

1. a. From 3 x 3 — 4 x + 7 x 2 — 11 subtract 2 x z + x + 
3 a: 2 + 3. 

b. What is the value of the remainder when x = 2 ? 

2. a. Find the product of 3 £ 2 — 4 x — 5 and 2 x — 3. 
b. Check the solution by letting x = 2. 

3. a. Simplify 3 m —(2 m — 7) + (6 — 4 m). 

b. What is the value of the result when m = — f ? 

4. Simplify 3 x —\x — 3{x — 1)}. 

5. Solve the equation 2 x — [x —(2 x — 5)] = 10. 

6. Is 4 a root of the equation x 2 — 5a: + 4 = 0? 

7. Divide xr 2 + 2 irrh by r. 

8. If S = tt 2 + 2 irrh, find S when x = - 2 /, r = 7, and A = 8. 

9. How much less than Ois — a; 2 + a:+l? 








92 


ALGEBRA 


73. Division of polynomials by polynomials is like long 
division in arithmetic. 


Example 1. Divide 864 by 24. 

Solution. 1. 86 -f- 24 = 3 + . 

2. 24 X 3 = 72; subtract . . 

3. 14 ^ 2 = 6 + . 

4. 24 X 6 = 144; subtract . . 


24 


36 

864 

72 

144 

144 


Example 2. Divide 10 x 3 — 21 x 2 — 11 x + 12 by 5 x — 3. 

2 z 2 — 3 x — 4 


Solution. 1. 10 x 3 + 5 x = 2 x 2 5 x — 3 

2. (5 x — 3) X (2 x 2 ); subtract 

3. (- 15 x 2 ) + (5 x) = — 3x 

4. (5 x — 3) X (— 3 x ); subtract 

5. (- 20x) + (5x) = - 4 

6. (5 x — 3) X (5 z); subtract 


10 x 3 - 21 x 2 - 11 x + 12 
10 x 3 — 6 x 2 

- 15 x 2 - 11 x 

— 15 x 2 -f 9 x _ 

- 20 x + 12 

- 20 x + 12 


Check. When x = 2, 

The divisor, 5 x — 3 = 10 — 3, or 7; 

The dividend, 10 z 3 - 21 z 2 - 11 x + 12 = 80 - 84 - 22 + 12, or - 14 
The quotient, 2 x 2 — 3 a: — 4=8 — 6 — 4, or — 2. 

Also (- 14) + (7) = - 2. 


Rule. — To divide a polynomial by a polynomial: 

1. Arrange the dividend and the divisor in either ascending 
or descending powers of some common letter. (See § 41, p. 45.) 

2. Divide the first term of the dividend by the first term of 
the divisor, and write the result as the first term of the quotient. 

3. Multiply the whole divisor by the first term of the quotient; 
write the product under the dividend and subtract it from the 
dividend. (Keep like terms in columns, as in Example 2.) 

4. Consider the remainder a new dividend, and repeat Steps 
2 and 3. (As in arithmetic, there may be a final remainder.) 

Historical Note. — Sir Isaac Newton pointed out the advantage 
of arranging the dividend and divisor in ascending or descending powers 
of the same letter. 










DIVISION 


93 


EXERCISE 53 

Divide: 

1. ar 2 -f 5 x + 6 by x + 3 

2. z 2 + 8 x + 12 by x + 2 

3. y 2 + 8 y + 15 by y + 5 

4. ra 2 + 13 m + 12 by ra + 1 

5. A 2 + 10 A + 24 by A + 6 

6. r 2 — 18 r + 32 by r — 2 

7. s 2 — 12 s -f- 35 by s — 7 

15. n 4 —5 n 2 — 24 by n 2 + 3 

16. x 2 — 16 ax + 55 a 2 by x — 5 a 

17. a 2 — 3 ab — 108 b 2 by a + 9 b 

18. ic 2 — 2 xz — 48 z 2 by x — 8 z 

19. x 4 + 6 x 2 y — 27 y 2 by $ — 3 y 

20. x 2 y 2 — 18 xy + 45 by xy — 3 

21. 15 x* — 14 x — 16 by 3 x + 2 

22. 6 a 2 + 25 — 25 a by 2 a — 5 

Suggestion. Read Step 1 of the rule, and § 41, page 45. 

23. 24 a 2 + 15 — 38 a by 6 a — 5 

24. 30 x 2 + 7 — 47 x by 6 x — 1 

25. 28 x 2 — 15 y 2 + 23 xy by 4 x + 5 y 

26. 20 m 2 + 41 mn + 20 n 2 by 5 m + 4 n 

27. z 3 — 5 x 2 - 17 x + 66 by x - 6 

28. 6 a 3 — 27 a - a 2 + 20 by 3 a - 5 

29. 4 x 3 - 5 xy 2 + 4 y z - 8 x 2 y by 2 x - y 

30. 12 a 3 + 17 a 2 b - 25 6 3 - 20 ab 2 by 3 a + 5 b 

31. 2 r* — x 2 y — xy 2 — 3 y 3 by x 2 + xy + y 2 

32. 5 n 3 + 8 n 2 — 23 n — 1 by n + 3 

33. 15 x z — 30 x — 8 — 19 x 2 by — 5 x + 3 x 2 — 4 


8. t 2 + 72 - 17 t by t - 9 

9. c 2 + 72 - 18 c by c - 6 

10. d 2 + 13 d + 36 by d + 4 

11. x 2 + 2 x — 8 by x — 2 

12. t 2 + t - 30 by t + 6 

13. a 2 — 6 a — 27 by a — 9 

14. to 2 + ra — 20 by ra + 5 

Check when n = 3. 

Check when x = 3, a = 2. 

{ Check by 
multiplication 

Check when x = 2, i/ = 4. 
Check when x = 2. 


94 


ALGEBRA 


74. In division of a polynomial by a polynomial, spaces should 
be left for any powers of the letter according to which the poly¬ 
nomial is arranged, which may not be present in the dividend; 
also there may be a remainder. 


Example. Divide x 2 y 2 + x 4 — y 4 by — xy + y 2 + x 2 . 
x 2 + xy -f y 2 


Solution. x 2 — xy + y 2 

x 4 + x 2 y 2 

x 4 — x z y + x 2 y 2 


y 4 


+ x 3 y 

+ x z y — x 2 y 2 

+ xy z 

y 4 


x 2 y 2 

- xy 3 - 

y 4 


x 2 y 2 

- xy 3 + 

z/ 4 


-2 y 4 

Note. — The quotient is x 2 + xy + y 2 ; the remainder is — 2 y 4 . 
As in arithmetic, the complete quotient may be written: 


x 2 + xy + y 2 + — 
x 2 


- 2 y 4 
- xy + y 2 


Divide: 


EXERCISE 54 


1. X 4 + x 2 y 2 + y 4 by x 2 + xy + y 2 


2. 16 n 4 — 81 by 2 n + 3 

3. y 3 + 27 by y + 3 

4. y 3 — 27 by y + 3 

5. x 3 + 2/ 3 by x — y 

10. 1 — y 5 by 1 

11. 27 m 3 + 64 n 

12. a 3 — b 3 by a + b 

13. 8 m 3 — 1 by 2 m + 1 

14. x 3 — 8 y 3 by x — 2 y 

15. 27 m 3 — 64 by 3 m + 4 

16. a 6 - b 3 by a 2 - b 

17. 1 - 8 t 3 by 1 - 2 t 

18. 1 + 27 c 6 by 1 - 3 c 2 

26. 16 a 8 - 


6. a 3 + b 3 by a + b 

7. 8 a 3 — 1 by 2 a — 1 

8. x 4 — 16 y 4 by x — 2 y 

9. x 6 * — 8 y 6 by x 2 — 2 y 2 
■ V 

by 3 m + 4 n 

19. x 4 — y 4 by x — y 

20. x 4 — y 4 by x + y 

21. x 5 — y 5 by x — y 

22. x 4 + 16 y 4 by x — 2 y 

23. a 8 + b 4 by a 2 - b 

24. 1 + 81 c 8 by 1 - 3 c 2 

25. x 4 — 81 y 4 by x — 3 y 
by 2 a 2 - 1 








VII. SIMPLE EQUATIONS 

75. An identity is an equation which is satisfied by all values 
of the literal numbers for which the two sides have any real 
meaning. 

Thus, 3 x(a - b) = 3 ax - 3 bx is an identity. 

If a = 3, b = 1, and z = 2, 3 x{a - b) = 3 • 2(3 - 1) = 6 • 2, or 12. 

Also, 3 ax — 3 bx = 3 • 3 • 2 — 3 • 1 • 2 = 18 — 6, or 12. 

Any other values of a, b, and x also satisfy this equation. 

Similarly, (3 a + 2 6) + (a — b) is identically equal to 4 a + b 
15 x 2 — 3 x 2 is identically equal to 12 x 2 
30 y z — 3 y is identically equal to 10 y 2 

That is, the result obtained from adding , subtracting, multiplying, or 
dividing two expressions is identically equal to the indicated combination 
of the numbers. 

76. A conditional equation is an equation involving one or 
more literal numbers, which is not satisfied by all values of the 
literal numbers. 

Thus: a. x + 2 = 5 is not satisfied by any value of x except x = 3. 

b. x 2 — 5 x = — 6 is satisfied by x = 2 and by x = 3, but by no 
other values of x. 

The word “ equation ” usually refers to a conditional equation. 

77. To solve a conditional equation is to find the numbers, 
called roots, which satisfy the equation. 

78. If an equation has only one unknown number, if the 
unknown does not appear in the denominator of any fraction, 
and if the unknown appears only with the exponent 1, then the 
equation is called an equation of the first degree, or a simple 
equation. 

Thus, S x — 5 = 4 is a simple equation. 

Historical Note. — The idea of the degree of an equation was 
introduced by Descartes. 


95 


96 


ALGEBRA 


79. There are three mechanical methods of solving a simple 
equation which can he used, although use of them is unnecessary. 


1 . 


80. Transposition. 

Example. Solve the equation 10 x — 5 
Solution. 

2. Ae 

3. S 32 ; 

4. C. T. 

5. D 7 


3 x + 30. 


10 x — 5 = 3 x + 30. 

10 x = 3 x + 30 + 5. 

10 x - 3 * = 30 + 5. 

7 x = 35. 
x = 5. 

By adding 5 to both members of Step 1, — 5 is made to disappear 
from the left member and + 5 to appear in the right member in Step 2. 
It is as if — 5, with its sign changed, were taken from the left side and 
placed on the right side. 

In Step 3, subtracting 3 x from both members of Step 2 causes 3 x 
to disappear from the right side of Step 2, and — 3 x to appear in the 
left side in Step 3. Again, it is as if 3 x, with its sign changed , were 
moved from the right side to the left side. 

These are two examples of transposition (“place across”)• 


Rule. — A term may be transposed from one member of an 
equation to the other, provided its sign is changed. 

Historical Note. — About the first quarter of the ninth century 
an Arabian mathematician, Mohammed ben Musa, wrote an algebra, 
for the title of which he used Ilm al-jabr wa’l muqabalah. Al-jabr 
meant the process of transposing terms. This title was used in various 
forms in Europe until about the fifteenth century, when the last part 
was dropped and algebra came into use. 


81. Canceling terms in an equation. 

Example. Solve the equation x + a = h + a. 

Solution. 1. x + a = b + a. 

2. S a x = b. 

The result is the same as if + a were simply dropped from both mem¬ 
bers. It was not simply dropped; it was subtracted from both members. 

Rule. — If the same term, preceded by the same sign, occurs 
in both members of an equation, it may be canceled (dropped) 
from both members. 


SIMPLE EQUATIONS 


97 


Example. Solve the equation 

5x + 9 x — 8+3 = 9 z + 18 — 8+2 z. 

Solution. 1. 5 x +J9-3T — $ + 3 =-9^r + 18 — $ + 2 x. 

2. Since + 9 x appears in both members, it can be canceled; also 
— 8 can be canceled. 5 x + 3 = 18 + 2 x. 

3. Transposing 

+ 2 x and + 3, 5 x — 2 x = 18 — 3. 

4. C. T. Sx= 15. 

5. D 3 x = 5. 

Check. Substitute 5 for x in the original equation. 

Note. — First, cancel equal terms which appear in both members; 
second, transpose terms containing unknowns to the left side, and 
terms containing knowns to the right side. 


EXERCISE 55 


Solve the following equations: 

1. 4m — 2 = ra + 7 

2. 8 t + 5 = 3 2 + 35 

3. 7 x — 10 = 50 — 5 x 

4 . 5 y -3 - 11 +2i/ 

6. 7 J2 — 11 = 15 — 6 2 


6. 15 r — 3=4 — 6 r 

7. 18# — l = 6x + 7 

8. 32y — 6 = 8 ?/ + 9 

9. 17 £ + 2 = 72 + 5 
10. 9m - 2 = 4m + 2 


11. 13 w - 92 + 8 = 4 + 5 w + 8 

12. 12 50 + 1= 80 + 3—5# 

13. 11a — 14 + a = 9a + 4 + a 

14. 20 5 — 19 ^ + 3 = - 19 5 + 13 - 5 s 

15. 4 x — 7 + 5# — 9 = 61+5# — 3 x 


16. 32 + + 16 + 14 + + 1 

17. 17 m + 16 m — 5 m — 4 

18. 14 2-13 + 112-17- 

19. 8 k + 6.5 -48 - 73 k = 

20. 50 r - 1.5 = 20 r + 1.8 

21. 20 z + 2.8 = 5 z + 6.7 

22. 7 w — 5.1 = 3 w + 2.5 

23. 6.25 + 3 = 4.75+18 


-9+ = 7+ + 16 + 14+ + 13 
= + 5m + 8—3w + 2 + 16m 
It = 82 - 13 - 162 + 25 + 11 2 
48 + 5 k + 6.5 — 17 k — 73 k 

24. 4.5a - 6 = 2.3 a + 11.6 

25. 3.5 g — 12.4 = 1.5 g + 6.8 

26. 7.5 x + 14.4 = 3.2 x + 48.8 

27. 15 y - 13.3 = 25.8 -Sy 


98 


ALGEBRA 


82. Changing signs in an equation. Often we obtain an 

equation like — x = + 3. 

We prefer that the sign of x be plus. To make it so, multiply 
both members by — 1. Then, x — — 3. 

The result of multiplying the original equation by — 1 is the 
same as if the signs of its terms were simply changed. 

Rule. — The signs of all terms of an equation may be changed, 
without destroying the equality. 

83. Negative roots of equations. 

Example. Solve the equation 

7 — 5 x — 9 a; = 15 — 9 x — 3 x. 

Solution. 1. 7 — 5 x —-9-ar = 15 —SHcr — 3 x. 

2. Canceling — 9 x, 7 — 5a;=15 — 3 a;. 

3. Transposing + 7 

and — 3x, Sx — 5x = 15—7. 

4. C. T. - 2 x = 8. 

5. D_ a x = — 4. (Check as usual.) 


EXERCISE 56 

Solve and check the following equations: 


6. 11 + 7 p = 3p - 13 

7. 9r — 5=4r + 5 

8. 23 - 7 z = — 15 2 + 15 

9. 16 + 3 w = — 2w + 6 
10. 19 — 16 y = 24 y — 1 


1. 2 — 3x = x - 8 

2. 4m-5 = 6m + l 

3. 13 — 6x = 3 — 11 x 

4. 7 t + 10 = 19 t - 2 

5. 12 — 5 n = n + 36 

11. 3(m + 1) - 9 = 5(2 m + 7) - 6 

12. 8 t - 5(4 1 + 2) = 3(5 - 3 1) - 1 

13. 8 — 3 x(x — 4) = 2 — a:(3 x — 15) 

14. 6(3 - 2 c) - 5 c = 5 c + 7(4 - 3 c) 

15. 3(3 x + 7) - 4 (6 x — 3) = 10 - 6 (3 x - 4) 

16. 10 r - (4 r + 3) = 3(8 r - 2) + 4(1 - 4 r) 


SIMPLE EQUATIONS 


99 


84. Solution of problems. Review Exercises 39 and 40. 

EXERCISE 57 

1. If 13 be added to a certain number, and the sum be 
multiplied by 2, the product is 60. What is the number ? 

2. Separate 28 into two parts such that 5 times the smaller 
exceeds 4 times the larger by 5. 

3. Separate 43 into two parts such that 2 times the larger, 
diminished by 4 times the smaller, equals 7 less than the larger. 

4. There are three consecutive integers such that 4 times 
the first, diminished by the second, equals 12 more than 2 times 
the third. What are they? 

5. What are the three sides of the triangle whose perimeter 
is 121 in., if the second side is twice the first, and the third side 
is 19 in. shorter than the second ? 

6. The Lincoln Memorial at Washington stands on a plat¬ 
form whose perimeter is 676 ft. The length of the platform 
exceeds its width by 70 ft. Find the length and the width. 

7. Separate 72.5 into two parts such that 3 times the larger, 
diminished by 4 times the smaller, equals 8 less than the larger. 

8. The age of A is now 4 times the age of B. Four years 
from now A’s age will be one year more than 3 times B’s age 
then. What are their present ages ? 

9. John is 4 years older than Charles. Eight years ago 
twice Charles’ age exceeded John’s age then by 1 year. How 
old is each now ? 

10. The treasurer of a club, after collecting dues, had $17.00 
in nickels, dimes, quarters, and half-dollars. The number of 
dimes was 3 times the number of nickels ; the number of quarters 
exceeded the number of dimes by 2; the number of half-dollars 
was 3 less than 5 times the number of nickels. How many 
coins of each kind did he have ? 


100 


ALGEBRA 


EXERCISE 58 

Interest Problems 

1. What is the simple interest on $500 at 6% for 1 year? 
for 3 years ? for n years ? 

2. What is the simple interest on P dollars at 6% for 1 
year ? for 4 years ? for t years ? 

3. What is the simple interest on $300 at r% for 1 year? 
for 3 years ? for t years ? 

4. What is the simple interest on P dollars at r% for 1 
year ? for 5 years ? for t years ? 

5. The total income from two investments is $135. The 
first sum is invested at 6% and the second sum at 7%. The total 
amount invested is $2000. Find the two sums invested. 


Solution. 1. Let s = the number of dollars invested at 6%. 



No. of Dollars 

Rate of Interest 

Annual Interest 

First sum 

8 

6% 

.06 s 

Second sum 

(2000 - s) 

7% 

.07(2000 - s) 


2. .*. .06 s + .07(2000 - s) = 135. 


6. One sum is invested at 6% and a second sum at 8%. 
The total income is $228. The second sum is 4 times as large 
as the first. What are the sums ? 

7. A man has one sum invested at 5%, and a second sum, 
$1500 larger than the first, invested at 7%. His total income 
from these sums is $321. How much has he invested at each rate ? 

8. The total of certain two investments is $2500, — one 
sum at 5%, and one at 6%. The annual interest from the 
former is $29 less than that from the latter. How much is 
invested at each rate ? 

9. A man has $6500 invested at 6% and a second sum in¬ 
vested at 7%. If the income from the second sum exceeds that 
from the first sum by $170, what is the second sum ? 







SIMPLE EQUATIONS 


101 


10. If $2500 is invested at 6% and $1500 at 7%, how much 
must be invested at 8% to make the total income 7% of the total 
amount invested ? 

Suggestion. 1. Let x = the no. of dollars to be invested at 8%. 


Sums 

No. op Dollars 

Rate of Interest 

Interest 

First 

2500 

6% 

$150 

Second 

1500 

7% 

$105 

Third 

X 

8% 

.08 x 

Total 

4000 + x 

7% 

.07(4000 + x) 


2. /. .07(4000 + x) = 150 + 105 + .08 x. 

(Complete the solution.) 


11 . $2500 is invested at 5%. How much must be invested 
at 8% to make the total income 6% of the total investment? 

12. If $1500 is invested at 5%, and $2000 at 6%, how much 
must be invested at 7% to make the total income 6% of the total 
amount invested ? 

13. If $2500 is invested at 6%, and $3500 at 5%, how much 
must be invested at 8% to make the total income 7% of the 
total amount invested ? 

14. If $3000 is invested at 6% and $1500 at 4%, how much 
must be invested at 5% to make the total income $300? 

15. A total of $4500 is invested in two companies, one sum 
at 7%, and the other at 8%. The interest on the latter exceeds 
that on the former by $60. How much is invested in each 
company ? 

16. One sum is invested at 6% and a second, which is twice 
as large as the first, is invested at 5%. The total interest is 
$448. How much is invested at each rate? 

17. The income on $4200 for (a; + 2) years at 5% exceeds the 
income on $3000 for x years at 4% by $870. Find x. 
















102 


ALGEBRA 


85. Distance, rate, and time problems. 

If a train travels 40 miles per hour for 6 hours, it goes 240 miles. 
Observe the rate (r), 40 miles per hour; the time (t), 6 hours; the 
distance (d), 240 miles. 

The distance is expressed as a number of units of length; as 
feet, rods, miles. 

The distance equals the rate multiplied by the time. (d = rt) 

In the illustration above, 240 = 6 X 40. 

The time is expressed as a number of units of time ; as minutes, 
hours, days, etc. 

The time equals the distance divided by the rate, {t = d -s- r) 

In the illustration above, 6 = 240 40. 

The rate is expressed as a number of units of length in the unit 
of time. 

The rate equals the distance divided by the time, (r =d + t) 

In the illustration above, 40 = 240 4- 6. 

EXERCISE 59 

1. How far does a train go in 9 hours if the rate is : 

a. 25 mi. per hr. ? b. r mi. per hr. ? c. (x + 2) mi. per hr. ? 

2. How far does a train go in h hours if the rate is : 

a. 35 mi. per hr. ? b. x mi. per hr. ? c. (r — 3) mi. per hr. ? 

3. How long does it take to go 150 miles if the rate is: 

a. 20 mi. per hr. ? c. 50 mi. per d. ? 

b. n mi. per hr. ? d. (r + 5) mi. per d. ? 

4. How long does it take to go N miles if the rate is : 

a. 30 mi. per hr. ? 6. (r + 3) mi. per hr. ? c. x mi. per d. ? 

5. At what rate is a man traveling who goes 300 miles in: 

a. 10 hr. ? b. h hr. ? c. (x + 4) hr. ? d. m d. ? 

6. What is the rate if an object moves D miles in: 

a. 12 hr. ? b. 3 d. ? c. t hr. ? d. k min. ? 

7. Give the rule for finding: a. the rate; b. the time ; 

c. the distance. 


SIMPLE EQUATIONS 


103 


8. One man travels 25 miles per hour for x hours, and a 
second man travels 20 miles per hour for 2 hours less time. 

a . How far does the first man travel ? 

b. How long does the second man travel ? 

c. How far does the second man travel ? 

d. Suppose the sum of the distances traveled by the two men 
is 100 miles. Write the equation expressing this fact. 

e. Solve the equation, finding the time x, and check the solu¬ 
tion by finding the distances in parts a and c and comparing 
their sum with 100. 

9. A and B travel toward each other from points which are 
300 miles apart, — A at the rate of 20 miles per hour, and B at 
the rate of 25 miles per hour. In how many hours will they 
meet, if they start at the same time ? 

Solution. 1. Since they start at the same time and travel until 
they meet, they travel the same number of hours. 


Let h = the number of hours they travel. 


Then for 

the time is 

the rate is 

the distance is 

one man 

h hours 

20 mi. per hr. 

20 h mi. 

the other man 

h hours 

25 mi. per hr. 

25 h mi. 


3. .*. 20 h + 25 h = 300. Since they started 300 mi. apart and travel 
until they meet. 

(Complete and check the solution.) 

10. Two men who are traveling in opposite directions at the 
rates of 18 and 22 miles per hour respectively, started at the 
same time from the same place. In how many hours will they 
be 250 miles apart ? 

11 . A and B started towards each other at the same time from 
points which are 150 miles apart, and met in 5 hours. If A 
traveled twice as rapidly as B, what were their rates ? 

12. A and B started towards each other at the same time 
from points which are 253 miles apart, and met in 5.5 hours. A 
traveled 6 miles more per hour than B. What were their rates ? 








104 


ALGEBRA 


13. A passenger and a freight train, whose rates are 30 miles 
and 18 miles per hour respectively, start toward each other at 
the same time from points which are 168 miles apart. How 
soon will they meet and how far will each have traveled ? 

14. A freight and a passenger train started towards each 
other at the same time from points which are 225 miles apart. 
If the passenger train’s rate was 30 miles per hour, and they 
met in 5 hr., what was the rate of the freight train? 

15. A and B started towards each other at the same time from 
points which are 240 miles apart, and met in 6 hours. If A 
traveled 12 miles per hour more than B, what were their rates ? 

16. Repeat Example 15, assuming A’s rate was 10 miles 
more per hour than B’s rate. 

17. A freight and a passenger train started in opposite direc¬ 
tions from the same place at the same time. Assume that the 
passenger train traveled twice as rapidly as the freight train. 
What were their rates if they were 315 miles apart in 7 hours ? 

18. Two automobiles started at the same time from the 
same point, in opposite directions. The first traveled 5 miles 
more per hour than the second. At the end of 8 hours they 
were 360 miles apart. What were their rates ? 

19. A and B started toward each other at the same time from 
points which are 252 miles apart. A traveled 4 miles more per 
hour than B. How fast did each travel if they met in 7 hours ? 

20. An automobile party is traveling 15 miles per hour. At 
what rate must a second party travel in order to overtake the first 
in 3 hours if it starts from the same place 1 hour after the first ? 

Hint. — If the second overtakes the first, how do their “distances” 
compare? Which party traveled the longer time and how long? 

21. One automobile party is traveling 15 miles per hour. 
In how many hours will a second party, traveling 25 miles per 
hour, overtake the first, if the second party starts from the 
same place 2 hours after the first party left ? 


SIMPLE EQUATIONS 


105 


REVIEW EXERCISE VII 

1. Add 2-1- a 2 — ab + 3 b 2 , a 2 + ab — f- b 2 , and fa 2 — 2 6 2 . 

2. From f z 3 - f x 2 y + £ zy 2 - i y 3 subtract f x 3 - f x 2 y 
+ i xy 2 - i y 3 . 

3. Solve and check the equation .5 x — 7 = .8 x — 14.5. 

4. a. Simplify 3 a; —[7 x + 2(5 — 2 x)— 3(4 x — 6)]. 

b. Simplify 2 m — [— m — 3(4 — m)+ 5(2 — 3 m)]. 

5. Multiply 3 a 2 — b 2 -f 2 a& by 2 a& — 3 b 2 + a 2 . 

6. a. Divide 6 a: 3 + 14 x - 17 x 2 - 3 by 3 z 2 - 4 x + 1. 

5. Check the solution by letting x = 2. 

7. a. How many tons of soft coal can be placed in a bin 8 
feet wide, 3 yards long, and 2 yards high, if one ton occupies 
38 cubic feet ? 

b. If T represents the number of tons of soft coal in a bin 
l feet long, w feet wide, and h feet high, write the formula express¬ 
ing T in terms of l, w, and h. 

c. If l and w have fixed values, how does T change if h is 
doubled ? 

8. Separate 307 into two parts such that the larger exceeds 
6 times the smaller by 13. 

9. Find three consecutive even numbers such that 4 times 
the smallest exceeds the sum of the other two by 30. 

10. A’s age is now twice B’s age. A J s age four years from 
now will be 3 times what B’s age was three years ago. What 
are their ages now? 

11 . The total interest from two investments is $909. The 
first sum, invested at 6%, exceeds the second sum, invested at 
7%, by $2800. What are the two sums ? 

12. A, traveling 20 miles per hour, starts at 6 a.m. towards 
B, who is 300 miles away. At 9 a.m., B, traveling at 25 miles 
per hour, sets out to meet A. At what time will they meet ? 


106 


ALGEBRA 


13. Simplify (x + 3 y)(2 x - y)— (4 x - y)(x + 5 y). 

14. Complete the following statement: 

a — (b — c + d) = a( ?)&( ?)c( ?)d. 

15. Inclose the last three terms of 16 x 4 — a 2 + 2 ab — b 2 in 
parentheses preceded by a minus sign. 

16. Find the sum, the difference, and the product of: 

a. 5(m 2 — n 2 ) b. ax 3 c. 2 x n 

— 3 (ra 2 — w 2 ) — bx 3 — 7 .T n 

17. $3000 is invested at 5% and $1000 at X%. If the total 
income is $220, find X. 

18. a. Express the perimeter and the area of the rectangle 
whose width is m feet and length n feet. 

b. How does the perimeter change when m increases and n 
remains unchanged ? 

19. a. In a certain state, a married man with an income of 
between $1200 and $2000 must pay an income tax of 1% of the 
remainder after subtracting (a) $1200 from his income, and 
( b ) $400 for each child in his family. Make a formula for his 
tax, T dollars, if he has n children and an income of I dollars. 

b. Find by this formula T when I = $1650 and n = 0. 

c. How does T change when n increases ? 

20. Four times John’s age three years ago is the same as 
twice his age five years hence. Find his age. 

21. If x is an even integer, what is : 

a. The next larger integer ? 

b. The next larger even integer ? 

c. The next smaller even integer? 

22. A seedsman wishes to mix enough oats, worth 1-|^ per 
lb., and corn, worth 2^<f per lb., with 200 lb. of wheat, worth 
2^ per lb., to make a mixture of 1000 lb. of chicken feed, worth 
2<f, per lb. How many pounds of oats and of corn should he 
take? 




VIII. SPECIAL PRODUCTS AND FACTORING 

86. To factor an algebraic expression is to find two or more 
expressions which will produce the given expression when they 
are multiplied together. 

87. A number (or expression) which has no factors except 
itself and unity is called a prime number; as, 3, a, and x + y. 

A monomial is expressed in terms of its prime factors thus: 

12 a?b 2 c = 2- 2- 3 -a-a-a-6-5-c. 

88. Squaring a monomial. 

Development. 1. What does x 2 mean? (xy) 2 *! (2 r 3 s) 2 ? 

2. Find each of the following squares by multiplication: 

a. (2 xy) 2 ; b. (3 a 2 b 2 ) 2 ; c. ( — 2 r 3 s 4 ) 2 ; d. (— -f to 2 ) 2 . 

3. Compare the exponent of each letter of the square with 
the exponent of that letter in the given monomial. 

Rule. — To square a monomial, square its numerical coeffi¬ 
cient, and multiply the result by each of the literal factors of 
the monomial, giving each letter twice its original exponent. 

Thus : ( — 5 x 2 y z ) 2 = 25 x*y G ; ( — § xy 2 ) 2 = -J x 2 y 4 . 

EXERCISE 60 

1. What sign does the square of any number have ? 

2. Learn thoroughly the squares of the integers from 1 to 20. 

Give at sight: 


3. 

(a 3 6) 2 

10. 

(+ 9 m 3 6 2 ) 2 

17. (-f abf 

4. 

(- a 2 6 3 ) 2 

11. 

(—11 rsff 

18. I 

/9x\ 2 

6. 

(3 a: 2 ) 2 

12. 

(14 mxy 3 ) 2 

\ 

Uly/ 

6. 

{ab 2 c) 2 

13. 

(- 12 a 3 6 4 ) 2 

19. | 

f 12 1\ 2 

7. 

(— 5 to) 2 

14. 

(—15 a?y) 2 

l 13 ) 

8. 

(- 4 xy) 2 

15. 

a « 2 ) 2 

20. | 

/U m\ 2 

9. 

(- 8 a 2 b) 2 

16. 

(- §mn) 2 

107 

ll5 x 2 ) 



108 


ALGEBRA 


89. The square root of a monomial. If an expression can 
be resolved into two equal factors, it is said to be a perfect 
square, and one of the factors is said to be its square root. 

Thus, 4 a 2 6 6 is equal to 2 ab z X 2 ab z ; hence it is a perfect square 
and 2 ab z is its square root. 

Note. — 4 a 2 b z is also equal to (— 2 ab z ) X (— 2 ab z ), so that — 2 ab 3 
is also a square root. In this chapter, only the positive square root will 
be considered. It is called the principal square root. 

The symbol for extracting the square root is the radical sign, 
y /; the vinculum is usually combined with it, V . 

D evelopment. 1. What sign does the square of any monomial 

have ? 

2. When squaring a monomial, what do you do with the 
exponents of the literal factors ? with the coefficient ? 

3. In finding the square root, then, what should you do with 
the exponents of the literal factors ? with the coefficient ? 

4. Find and test the result by multiplication: 

a. Va?; b. V4 x 2 y 2 ; c. Vl6 rV; d. V25 x 2 y 2 z 6 . 

Rule. — 1. A perfect square monomial is positive, has a 
perfect square numerical coefficient, and only even numbers 
as exponents. 

2. To find its square root: find the square root of its nu¬ 
merical coefficient, and multiply the result by the literal factors 
of the monomial, giving each letter one half its original exponent. 

EXERCISE 61 


Find the indicated square root: 


1. V4 x 2 

2. V 16 y 4 

3. vW 6 

4. V25 a 2 b 4 


5. \/64 r 2 s 6 

6. V81 c 4 d 2 

7. V49 ~ct?b 6 

8. VlOO m* 


9. Vl69 a 2 b 2 

10. V225 c 6 

11. Vl96 x 4 

12. a/256 y 2 
















SPECIAL PRODUCTS AND FACTORING 


109 


13. V49 t 6 

14. Vi c 2 d? 

16 - V-| m 2 

16. Vif a 2 b 2 

17. x 4 


18. a/ 256 mV 

19. a 4 fe 6 

20. V^c 4 d 2 

21. Vl6 x*“ 

22. VlOOa 2 ^ 


23. V-^x 4 ” 

24. V t |-9 x 4n y 2 m 

25. Vfjfa 2a: & 4a; 

26. 

27. Vfffx 2 y* 


90. Examples involving decimals. 

Since .3 X .3 = .3 2 = .09, then (.3 x) 2 = .09 x 2 . 
Similarly (.4 m 2 ) 2 = .16 m 4 . Also V.0016 £ 4 = .04 £ 2 , 


Find mentally: 

1 . (.2 a ) 2 

2. (.4 rs) 2 

3. (.It 2 ) 2 

4. (.5 x 3 ) 2 

5. (.9 m 4 ) 2 


EXERCISE 62 

6. (.01 y) 2 

7. V.36 c i 

8. V.64 r 8 

9. V.04 x 2 
10. V.49 m 4 


11. V.09 c 6 

12. V.16 m 2 

13. (.05 t 2 ) 2 

14. V.0036 m 4 n 2 

15. (.07 c 3 d) 2 


FACTORING POLYNOMIALS 

91. It is not always possible to factor a polynomial. Fac¬ 
torable polynomials are the products of certain special forms of 
number expressions. 

Case I 

Type Form: a(b + c)= ab + ac 


EXERCISE 63 


Find mentally: 

1. 2 a(x — y + z) 

2. -3 x(x 2 + xy — y 2 ) 

3. + 4 cd(c 2 — cd + d 2 ) 

4. — \ x(4 x 2 — 6 x — 8) 

6. i mn( 12 m 3 — 3 m 2 n — 6 ti 3 ) 


6. \ r(a + b + c + d) 

7. -2 x°(3 x 2a — 4 x a + 5) 

8. -6 xy(x + y - 1) 

9. .5(3 a — 4 b + 6 c) 

10 . - .3(10 x - 8 y + 5 z) 

























110 


ALGEBRA 


92. Factoring a polynomial whose terms have a common 
monomial factor. 

Example. Factor 14 xy 3 — 70 x z y 2 . 

Solution. 1. 14 and 70 can both be divided by 2, 7, and 14. 

14 is the greatest common factor of 14 and 70. 

We select it as part of the monomial factor. 

2. rc is a factor of both 14 xy 3 and 70 x 3 y 2 . It is the highest power of 
x which is a common factor of 14 xy 3 and 70 x 3 y 2 . 

We select x as part of the monomial factor. 

3. y 3 appears in 14 xy 3 , but only y 2 appears in 70 x 3 y 2 . Then y 2 is 
the highest power of y which is a factor of both 14 xy 3 and 70 jfy 2 . 

We select y 2 as part of the monomial factor. 

4. .*. the monomial factor is 14 xy 2 . 

5. 14 xy 3 -v* 14 xy 2 = y ; — 70 x?y 2 -r- 14 xyf == — 5 x 2 . 

6. .*. 14 xy 3 — 70 x 3 y 2 = 14 xy 2 (y — 5 x 2 ). 

Note. — Do Steps 1, 2, and 3 mentally. 

Rule. — To factor a polynomial whose terms have a common, 
monomial factor: 

1. Find the greatest common factor of its terms. 

2. Divide the polynomial by it. 

3. The factors are the common factor found in Step 1, and 
the quotient obtained in Step 2. Check by multiplying. 

EXERCISE 64 


Factor the following polynomials : 


1. 

5 a 

+ 56 

6. 

P + 

Prt 

11. 

cd + cdx 

2. 

7 t 

- 7 r 

7. 

3 ax 2 

- 

3 ay 2 

12. 

m 2 n — mn 

3. 

3s 

- 12 x 

8. 

y na 

+ 

y wl 

13. 

6 a 4 — 12 a 2 h 

4. 

mx 

— my 

9. 

2 7rr 2 

+ 

2 7 rrh 

14. 

15 cd? — 5 cd 4 

5. 

7r r 2 

+ 7r rl 

10. 

5 ami‘ 

) 

- 15 an 2 

15. 

14 xy 2 — 2 x 3 

16. 

5 m 

3 — 10 m 2 

+ 15 

m 


20. 3 x 2 

+ 3 

xy + 3 y 2 

17. 

3 x 2n - 6 x n y n 

+ 3 y 2n 


21. 2 rV - 

4 r 2 z + 2 r 2 

18. 

ma 2 * + 6 ma x 

+ 9 m 


22. 6x 2 ° 

^ _ ' 

12 x Za -r 18 x 4a 

19. 

t3? ■ 

- 4 txy + 

4 ttf 



23. an 2 

- 7 

an — 18 a 


SPECIAL PRODUCTS AND FACTORING 


111 


24. 

1 

¥ 

rB + 

i s B 4~ 

its 

35. 

7r r 2 h + -J. 7 rR 2 h + J irrRh 

25. 

1 

¥ 

hci T 

hb + f hm 

36. 

at + 

igt 2 

26. 

9 

m 2 x a 

— 6 mx a 

- 15 x a 

37. 

2 -.2 . 
¥ x 

-if -I2 2 

27. 

7 

y 2 z n + 21 yz n 

- 126 

38. 

x s y - 

- x 2 y 2 — xy 

28. 

3 

ab 2x - 

- 15 ab x 

+ 18 a 

39. 

.5 m 

— .5 n + .5 p 

29. 

2 

mV 

+ 4 m?n 

— 126 m 3 

40. 

3z 2w 

- 3 y 2m - 6 z 2m 

30. 

4 

m 2 r 2 + 8 m 2 r 

- 60 m 2 

41. 

.4 r - 

- .8 s — .4 t 

31. 

24 m 2 a 

+ 18 mna — 15 n 2 a 

42. 

.2 w 

— .6 m — .8 t 

32. 

9 

cV - 

- 4 <?xy 

- 13 c s y 2 

43. 

\x 2 • 

- %xy + fxz 

33. 

3 

c 2 x a - 

-132 cx° 

- 135 x a 

44. 

.9 a - 

— .6 b + .3 c 

34. 

1 

2 

ax + 

| am + 

\ az — \ at 

45. 

.5 r - 

- 25 + 1.5* 


46. Suppose that the polygon ABCDEF 
can be divided into six triangles, such that 
their altitudes are equal. Call the altitudes 
each a, and the bases b, c, d, e, f, and g. 

a. What is the area of A OBC ? A ODC ? 

A ODE? etc. 

b. Indicate the sum of these areas. 

c. Simplify that sum by removing the monomial factor. 

d. Simplify the result by substituting p for (6 + c + d + e 
+ / + 9 )’ 

47. Suppose that the altitude of A RX T is a, and the alti¬ 
tude of A RST is c. The base of each is b. t 

a. Represent the area of each. 

b. Indicate the sum of these areas. 

c. Simplify the result. x- 

48. The area of a circle whose radius is rr is 7rr 2 . 

a. What is the area of the circle of radius R ? 

b. Indicate the area of the ring between the 
large and small circles. 

c. Simplify the result. 










112 


ALGEBRA 


Case II 


Type Form: (a + b)(a — b) = a 2 — b 2 

93. The product of the sum and the difference of two numbers. 

Note. — By “difference of” two numbers, we mean the first minus 
the second. 

Development. 1. Find by multiplication as in § 58 the 
following products, and write the results as in part a: 

d. {x + 3)0 - 3) = z 2 - 9 c. 0 + 10)0 - 10) = ? 

b. (m + 7)(m — 7) = ? d. (r + 9)(r — 9) = ? 

2. Observe the results in Step 1; try to find the following 
products mentally. Check the results by multiplication. 

a. ( a + 6)(a — 6).= ? c. (d + 4 )(d — 4) = ? 

b. (c + 8)(c - 8) = ? d. (y + 5)0 - 5) = ? 

3. Find by multiplication ( a -f- b){a — b). 

Rule. — To find the product of the sum and the difference of 
two numbers: 

1. Square each of the numbers. 

2. Subtract the second square from the first. 

Example. Find (5 a 2 + m)(5 a 2 — m). 

Solution. (5 a 2 + m) (5 a 2 — m) = (5 a 2 ) 2 — (m) 2 , or 25 a 4 — m 2 . 

Note. — When doing such multiplication, do not write (5 a 2 ) 2 — (m) 2 . 
Do that part mentally and give at once the result 25 a 4 — m 2 . 

This is called finding the result by inspection. 


EXERCISE 65 


Find by inspection: 

1. (x + 3)(z - 3) 

2. (m + 5)(m — 5) 

3. (t 2 - 2 )(t 2 + 2) 

4. ( x 2 — 6)(z 2 + 6) 
6. (m 3 -b 4) (m 3 — 4) 


11. (10 2 + w 3 )(10 s — u?) 

12 . (4 m — .5 n) (4 m -f .5 n) 


7. (r 2 + 3s)(r 2 -3 s) 

8. (3 x — b) (3 x + b) 

9. (7 m - 4)(7m + 4) 
10. (5a: 2 - y)(y + 5 z 2 ) 


6. (a + 2 6)(a —2 b) 


SPECIAL PRODUCTS AND FACTORING 113 


13. 

(6 x 2n — 7 y a ) (6 x 2n + 7 y a ) 

24. 

(i< 2 

+ 

iU 

1 

14. 

(ab — c)(c + ab) 

25. 

(4* 

- *)(** + *) 

16. 

(a 5 — 6 5 )(a 5 + b 5 ) 

26. 

(| a 2 

-|6)(|a 2 +f 6) 

16. 

(x 2a + y 2b )(x 2a — y 2b ) 

27. 

(10 - 

- x 2 )^ 2 + 10) 

17. 

(a 4 + 6 4 ) (a 4 - V) 

28. 

(c 2 + 

12) (12 - c 2 ) 

18. 

(a + b)(a — b)(a 2 + b 2 ) 

29. 

(1 - 

4 xi/) (1 + 4x?/) 

19. 

(x s + y 3 )(x 3 - y*)(x* + 2 / 6 ) 

30. 

(11 m Zx + 1) (11 m 3x — 1) 

20. 

(3 xy 2 — 5 2 s ) (3 xy 2 + 5 2 3 ) 

31. 

(i^ 

— 1)(^ x + 1) 

21. 

(i * a ~ i V b )(i x a + i y h ) 

32. 

(.5 a 

+ 2) (.5 a - 2) 

22. 

i)(f m-i) 

33. 

(.3 n 

— .2 r)(.3 n + .2 r) 

23. 

(ln 2 -i)(in 2 + f) 

34. 

(.6 c 2 

- 5) (.6 c 2 + 5) 

36. Simplify (x — 5)(x + 5) 

- 0 

+ x)(3 - x). 


Solution. 

(x - 5) (x + 5) - (3 + x)(3 - x) = (x 2 - 25) - (9 - x 2 ). 
(Complete the solution by removing parentheses and combining terms.) 

Note. — Observe that the product of (3 + x) and (3 — x) is 9 — x 2 
and that this product is placed in parentheses. Always do this if you 
would avoid errors in signs. 

36. Simplify (5 a — 2)(5 a + 2)— (3 a + 1)(3 a — 1). 

37. Simplify (x — y)(x + y)+(2 x — y)(2 x + y). 

38. Simplify (1 — 3 ic)(l + 3 x)— (1 — 4 x)(l + 4 x). 

39. Simplify (2 a— b) (2 a + b) — (a — 2 b) (a + 2 b). 

40. Simplify (m + 3 n)(m — 3 n)— (2 n + m)(2 n — m). 

41. Simplify (e — -J- b)(c + b)— (■$• c + 6)(J- c — b). 

42. Find mentally the product of 24 and 16. 

Solution. 1. 24 X 16 = (20 + 4) (20 - 4) 

2. = 400 - 16, or 384. 

Find mentally: 

43. 18 X 22 46. 25 X 15 49. 33 X 27 62. 42 X 38 

44. 17 X 23 47. 53 X 47 60. 44 X 36 63. 28 X 32 

46. 41 X 39 48. 34 X 26 51. 31 X 29 64. 37 X 43 


114 


ALGEBRA 


94. Factoring the difference of two perfect squares. 

Development. 1. What is the product of (a + 2) and 
(a — 2) ? What, then, are the factors of a 2 — 4 ? 

2. Find the factors of: 

a. a 2 — 9 b. m 2 — 16 c. k 2 — l 2 d. 9 r 2 — 4 s 2 

Rule. — To factor the difference of two squares: 

1. Find the square roots of the two perfect square terms. 

2. One factor is the sum of the results; the other factor is 
the difference of the results. 

Example. Find the factors of 25 r 4 — 16 t Q . 

Solution. 1. 25 r 4 — 16 t e = (Sr 2 ) 2 — (4 if 3 ) 2 . 

2. 25 r 4 -16 Z 6 = (5 r 2 + 4 1?)(5 r 2 - 4 Z 3 ). 

Note. — Examine the given expression carefully to make certain 
that it is the difference of two squares, as in Step 1. 

However, when giving the factors, omit Step 1, and at once write 
what appears in Step 2. 


EXERCISE 66 

Factor the following when possible: 


i. 

x 2 - 

64 

12. 

49 - y 

,6 

23. 

64 x 2 y^ - 

-2 6 

2. 

y 2 - 

144 

13. 

196 - 

a 2 

24. 

81 t 2x 

— 

169 m 4y 

3. 

100 ■ 

-2 2 

14. 

256 c 2 

- d 2 

25. 

100 aV 

- c 4 

4. 

z 4 - 

y 2 

15. 

16 x 2 - 

- 50 y 2 

26. 

25 „2 
T9 a 

- 

3 6 7,2 

T 2 T 0 

5. 

9 m 2 

- l 

16. 

~2 _ l 

x T 


27. 

225 w? x - 

-256 n 2 y 

6. 

x* - 

y b 

17. 

9 c 2 - 

i fp 

16 a 

28. 

36 

169 

7. 

x 2 - 

16 y 2 

18. 

1 ™2 
25 m 

- 1 

x 2 

y 2 


8. 

9. 

a 2 b 2 
36 t 6 

- 25 

- 1 

19. 

20. 

4 ~.2 _ 
"9 X 

144 x 8 

49 

- y 6 

29. 

a 4 

25 

fc 4 

49 


10. 

81 m 

2 - a 4 

21. 

9 ™2m 

- l 

30. 

16 x 2 


TO 2 

11. 

121 a b - 16 b 2 

22. 

196 a 2 

- fc 4 

y 2 


25 


31. Find mentally the value of 13 2 — 7 2 . 
Solution. 13 2 - 7 2 = (13 + 7) (13 - 7) 

= 20 X 6, or 120. 



SPECIAL PRODUCTS AND FACTORING 115 


Find the values of the following mentally, if possible. 

32. 17 2 - 8 2 3 5. 34 2 - 6 2 3 8. 33 2 - 27 2 

33. 14 2 - ll 2 36. 26 2 - 14 2 39. 46 2 - 40 2 

34. 22 2 - 8 2 3 7. 21 2 - 19 2 40. 27 2 - 23 2 

41. Factor (m — n ) 2 — 25 a 2 . 

Solution. Just as x 2 — y 2 = (x + y) — y) 

so (m — nY —25 a 2 — {(m — n) — 5 a}{(m — n) + 5 a} 
= {m — n — 5 a}{m — n + 5 a}. 

Factor the following: 


42. (a - b) 2 - c 2 

43. x 2 — (z — y ) 2 

44. (2 a — b) 2 — 9 c 2 

45. 16 s 2 — (t + w) 2 


46. 25 a 2 - (2 - 3 a) 2 

47. (2 x - 5) 2 - 16 

48. 49 - (3 + x) 2 

49. 16(c — d) 2 — 9(x — y ) 2 


Factor if possible: 

50. 9x 2 - 16 y 2 

51. 25 r 4 - 81 

52. 36 c 2 - 49 d 6 

53. 100 m 4 -169 

54. 196 a 2x - 256 


55. 400 x* n - 

56. - 25 m 2 

57. -fo a 2 - 4 b 2 

58. c 8 - 36 m 

59. 9 c 5 - 16 


60. .25 a 2 - 1 

61. 9 b 2 — .16 

62. .49 x 4 — .04 

63. .09 t 2y - .36 

64. .64 c 4 - .81 


, Miscellaneous Review 

Find: 

65. 5 c(3 m + 4 n — 8 p) 

66 . (2 xy — 3 z) (2 .ry + 3 z) 


67. 33 X 47 

68. (.2 a + .5 6) (.2 a — .5 b) 
Factor: 

73. 3 x s y 2 + 6 £ 2 ?/ 2 + a:?/ 2 + 2 y‘ 

74. 225 m 4 - 256 n 2 

75. 9 a 2 - .25 

76. 14 c s d + 10 c 2 d — 2 cd 

77. f X 17 + f X 13 


69. - f*(5 - 10® + 15 x 2 ) 

70. .6 tf(4 a — 7 6 -h *3 c) 

71. 82 X 98 

72. (f - .05) (f w a + .05) 

78. 1.44 z 2 — 25 y 2 

79. | X 21 + f X 19 

80. 400 x 6a — 900 y 2b 

81. 

82. .09 m 2 — .49 ?i 2 


116 


ALGEBRA 


95. Complete factoring. When a number or an expression 
is to be factored, all its prime factors should be found. Often 
factors selected at first may be factored again. 

Example 1. 48 = 8 • 6 

= 4-2-2-3 
= 2 • 2 • 2 • 2 • 3. 

Example 2. Factor 36 ty 8 — 36 tz 8 completely. 

Solution. 1. 36 ty 8 - 36 tz 8 = 36 t(y 8 - z 8 ) 

2. = 36 t(y 4 + z 4 )(y 4 - z 4 ) 

3. =36% 4 + zW + zW-* 2 ) 

4. = 36 t(y 4 + z 4 )(?/ 2 + z 2 )( 2 / + z)(y — z). 

Rule. — To find the prime factors of an expression : 

1. First remove any monomial factor which may be present. 

2. Then factor the resulting expression, when possible, re¬ 
writing all expressions which cannot be factored. 

Note 1. — Since only a few elementary cases in factoring are being 
taught in this chapter, you may at times get factors which are not 
factorable by any methods you know. “Factor completely,” there¬ 
fore, means factor as far as possible, using the methods taught in this 
chapter. 

Note 2. — You probably think that x 2 + y 1 ought to be factorable. 
It is not. Neither is x 4 + y 4 . 


EXERCISE 67 

Factor as completely as you can the following: 


1. 6 m 2 — 6 n 2 

2. ac 2 — acP 

3. 4 r 2 — 4 s 4 

4. 5 am 2 — 20 an 2 
6. 16 x 2 y — 25 yz 2 

6. f* 2 -f y 4 

7. 75 a 2 - 12 b 2 

8. 81 xt 2 — 49 xy 2 


9. 32 m 4 - 18 n 2 

10. 7rr 2 — 7 rs 2 

11 . x 8 z — y A z 

12. 4 x 4a - 4 y 4a 

13. 16 at 4 — 16 am 4 

14. x 12 — y 12 

15. 5 x 8 — 5 y 8 

16. 7 am 4 — 7 an 5 


17. J ax 4 — \ ay 4 

18. f cm 6 — -J cn 6 

19. f ar 8 — ^ as 8 

20. f xy z — f £3 6 

21. .5 x 2c - .5 y 2e 

22. .15 a 4 - .15 b 4 

23. .6 x 4 + .6 y 4 

24. .03 m 8 — .03 n 8 


SPECIAL PRODUCTS AND FACTORING 


117 


Case III 


Type Form : (ax + b) (cx + d) 

(See the Note with § 97, on page 120.) 

96. The product of two binomials of the form (mx -f- p ). 


Development. 1. Below, 
(2 x + 3 y). 

5 x — 4 y 
2x + 3 y 
10 x 2 — 8 xy 

+ 1 5xy -Yly 1 - 

10 x 2 + 7 xy — 12 y 2 


(5 x — 4 y) is multiplied by 


- 8 xy 
+ 15 xy 
+ 7 xy 


Observe the 
products 
which give 
the middle 
term. 


2. Similarly find the following products, and write the result * 
thus: 

(5 x — 4 y)(2 x + 3 y) = 10 x 2 + 7 xy — 12 y 2 

a. (2z + 3)(3x + l) c. (4 ?/ — 3) (2 y — 1) 

b. (3m+2)(2m + 3) d. (3 y + 2)(2 y — 4) 

3. Examine carefully the results in Step 2. Then try to get 
the first and third terms of the following products mentally. 
Possibly you can also get the middle term mentally. Check 
by long multiplication. 

a. (2 r + 4)(3 r + 2) c. (2s + 3)(2 5 + 1) 

b. (3 r + 5) (r + 2) d. (4 5 + 2) (2 s + S) 

4. In all the examples of Step 3, the only difficulty is that of 
getting the “ middle term.” In Step 1, this middle term, 

+ 7 xy, is the sum of — 8 xy and + 15 xy. — 8 xy is the 
product of 2 z and —4 y; + 15 xy is the product of 5 a; and 
3 y. These are called the cross products, because of the manner 
in which the arrows cross each other. 

Observe t he locatio n of these cross products below. 

1. (5 ^Ty^^TSy) = 10 x 2 + (- 8 xy + 15 xy) - 12 y\ 

2. /. (5 x — 4 y ){2 x + 3 y) = 10 x 2 + 7 xy - 12 y\ 

When finding such products, write only the second line. 







118 


ALGEBRA 


Rule. — 1. The first term of the product is the product of the 
first terms of the binomials. [5 x • 2 x = 10 x 2 .] 

2. The middle term of the product is the algebraic sum of 
the cross products. 

[In the example on page 117, follow the curved lines: 

(- 4 y) • (2 x) = - 8 xy, and (5 x) • ( + 3 y) = + 15 xy; 

(- 8 xy) + (+ 15 xy) = + 7 xy.] 

3. The third term of the product is the product of the second 
terms of the binomials. [(—4y)*(+3y)= — 12 y 2 .] 

Example. Find (5 r — 6 5 ) (2 r + 3 s). 

Solution. (5 r — 6 s) (2 r + 3 s) = 10 r 2 + 3 rs — 18 s 2 . 

Note. — Only the line above should be written. The thinking, 
which produces it, is: 

a. First term : 5 r • 2 r = 10 r 2 . 

b. Second term: — 6 s • 2 r — — 12 rs; 5 r • 3 s = + 15 rs; 

(- 12 rs) + (+ 15 rs) = + 3 rs. 

c. Third term: — 6 s • + 3 s = — 18 s 2 . 

EXERCj.SE 68 

Find mentally: 


1. 

(x 

+ 

l)(x + 2) 

14. (6 b - 

- 5)(2 b + 1) 

2. 

(x 

+ 

2)(x + 3) 

15. (4 m 

- 7)(2m + 3) 

3. 

(x 

+ 

4)(z + 2) 

16. (5 t - 

- 8)(3 ( + 2) 

4. 

(3 

x ■ 

+ 1)(* + 3) 

17. (x - 

7)0 + 9) 

6. 

(3 

x ■ 

+ 1)(2 x + 3) 

18. (y + 

11)0 + 6) 

6. 

(x 

- 

4)0 - 1) 

19. 0 - 

14)(z + 10) 

7. 

(x 

- 

3)0 - 5) 

20. (w + 

20)(w - 5) 

8. 

(2 

X 

- 3) (2 x - 5) 

21. (3 m 

+ 2)(3 m + 2) 

9. 

(3 

X 

- 2)0 - 6) 

22. (5 m 

- 3) (5 m - 3) 

10. 

(2 

X 

- 3) (6 x - 1) 

23. (6 c - 

f 5) (6 c + 5) 

11. 

(a 

+ 

4)0 - 1) 

24. (10 k 

rP 

1 

o 

t-H 

iP 

1 

12. 

(a 

+ 

5)0 - 1) 

25. (12 m + 5)(12 m - 5) 

13. 

(a 

+ 

6) (a — 2) 

26. (6 ab 

— 5) (4 ab -f 1) 


SPECIAL PRODUCTS AND FACTORING 119 


27. (7 mn — 3) (5 mn — 2) 

28. (9 r 2 - 2)(8r 2 + 3) 

29. (13 a 2 + 6)(5x 2 - 2) 

30. (7 p 2 - 8) (4 p 2 + 5) 

31. (10 ri? + 3) (12 w 3 - 1) 

32. (9c -2 d)(5 c + 3 d) 

33. (11 x - 5 y) (2 x + ?/) 

34. (8 a — 11 b) (3 a + 4 6) 

35. (7m + 9s)(3m -5 5 ) 

36. (9 £ — 4 x)(8 t — 2 x) 

37. (2 ab — 7 c)(3 ab + 5 c) 

38. (10 mn 2 + 7) (4 mw 2 — 3) 

39. (5 + x)(2 + x) 

40. (3 y — 2 z) (4 y — 5 z ) 

41. (9 + 4 z 2 )(10 - 5z 2 ) 

42. (11 — 6 xy)(5 + 6 xy) 

43. (12 — 5 mn )(4 — 3 mn) 

44. (12 a 2 6 — 2 c)(a 2 6 — c) 

45. (12 x 2 + 11 2/)(2 x 2 — y) 

46. (8 t 2 - x 2 ) (6 t 2 - x 2 ) 

47. (11 m 2 — 4)(11 m 2 + 4) 

48. (x 2 — 6 y)(x 2 + 13 y) 

49. (12 a - 11 c)(12 a + 11 c) 

50. (15 to — 4 n)(9*m + n) 

51. (19 a — b) (3 a — b) 

52. (9 c 3 - 2d)(7c 3 + d) 

53. (x 2 — 7 ?/)(x 2 — 10 y) 

54. (11 to 2 — 5 w)(4 to 2 + 5 n) 

55. (6p 2 - 7) (4 p 2 + 5) 

66. (13 x 2 + 7z)(5x 2 — 2 z) 


57. (x - 8?/)(|x + ?/) 

68. (6a - J6)(6a 

59. (ix — 8)(Jx + 9) 

60. (f a -i)(12a - 15) 

61. (\x - %y)(6 x +8?/) 

62. (6 x — 7 1 /) (6 x + 7 ?/) 

63. (3 to + 5 ft)(3 m + 5 n) 

64. (6 - 5t )(2 +40 

65. (1 - llz)(l + 12 z) 

66. (f- x — 2) (8 x + 1) 

67. (.2 x - .3) (4 x - 3) 

68. (.5 a + 2) (.3 a - 1) 

69. (.7 c - 5) (.2 c + 3) 

70. (8 x — .9 y)(2 x + .1 y) 

71. (19 a — 7 b)( 3 a — b) 

72. (5 — 3 x 2 )(2 + x 2 ) 

73. (St 2 — 5 x 2 ) (6 £ 2 — x 2 ) 

74. (5 a 3 + 6) (3 a 3 - 2) 

75. (4 b 2 + 5)(2 6 2 — 1) 

76. (3 c 3 + 4)(2 c 3 — 1) 

77. (x a + 7)(x° + 9) 

78. (r-3)(r-5) 

79. (x n + 2)(x n —5) 

80. (2 x c — l)(x c + 2) 

81. (3^+W- 1) 

82. (4 + x m ) (5 — x TO ) 

83. (7 — y m ) (3 + y m ) 

84. (3 + a x ) (2 — a x ) 

85. (x*-7)(2x»+ll) 

86. (2x a — 1)(3 x° + 1) 


120 


ALGEBRA 


97.* Factoring trinomials of the form x 2 + px +- q- 

Note. — This topic in factoring is optional in the Regents’ 1928 
Syllabus. It is part of the requirements of the College Entrance 
Examination Board. (Document No. 107, 1923.) 

Example. Factor x 2 + 7 x + 12. 

Solution. 1. This trinomial is the product of two binomials having 
x as their first terms. Place x in parentheses thus: 

(x ) (x ). 

2. The second terms may be 6 and 2, since 6 X 2 = 12. Place them 
in the parentheses, preceded by plus signs, since 7 x and 12 are both 
positive terms. Thus: (x + 6)(x + 2). 

This product has + 8 x as its middle term. Therefore these factors 
are incorrect. 

3. The second terms may be 3 and 4, since 3 X 4 = 12. Place them 
in the parentheses, preceded by plus signs. Thus: 

(x +- S)(x + 4). 

This product has + 7 x as its middle term, as it should be. These 
are the correct factors, since their product is x 2 + 7 x +- 12. 


EXERCISE 69 

Factor the following trinomials : 

1. z 2 + 8 a; + 12 5. t 2 + 10 t + 16 

2 . z 2 + 10 a: +- 24 6. y 2 + 12 y + 27 

3. r 2 + 11 r + 24 7. z 2 + 14 z + 40 

4. s 2 + 11 8 + 30 8. m 2 + 14 to + 33 

9. p 2 - 7 p + 12 

Hint. — Since 12 is positive, the second terms piust have like 
and, since the middle term is negative, these second terms, having like 
signs, must be negative. (Complete the solution.) 

10. x 2 - 7 x + 10 15. m 2 - 9 m + 14 

11. y 2 - 10 y + 21 16. r 2 - 12 r + 32 

12. z 2 - 92 + 20 17. s 2 - 10 s + 9 

13. a 2 - 12 a + 35 18. c 2 - 12 c +- 35 

14. b 2 - 9 b + 18 19. a 2 - 11 a + 30 


SPECIAL PRODUCTS AND FACTORING 121 


20 . Factor x 2 — 26 x — 192. 


Solution. 1. Since — 192 is negative, the second terms must have 
unlike signs. Since — 26 is negative, and is the sum of the second 
terms, the term having the larger absolute value must be negative. 

2. Since 192 is a large number, systematically list its factors. 

( + 1)(— 192); (+ 1) + (— 192) = — 191. Unsatisfactory. 

(+ 2)(- 96); (+ 2) + (- 96) = - 94. Unsatisfactory. 

( + 3)(— 64); (+ 3) + (- 64) = - 61. Unsatisfactory. 

(+ 4)( — 48) ; (+ 4) + (— 48) = — 44. Unsatisfactory. 

( + 6)(— 32) ; (+ 6) + (— 32) = — 26. These are satisfactory. 

3. .*. x 2 - 26 x - 192 = (x + 6)(x - 32). 

Note. — Step 2 should be done mentally. 


21. z 2 + 5 z — 14 

33. 

c 2 + 15cd + 36d 2 

22. w 2 — 3 w — 18 

34. 

x 2 y 2 — xy — 6 

23. t 2 - 3 t - 28 

35. 

m 2 — 7 m — 44 

24. m 2 — 4m — 45 

36. 

x 2a + 12 x a + 27 

25. r 2 — r — 42 

37. 

m 2x + 9 m x + 20 

26. s 2 + 6 5 - 27 

38. 

X 2c + 7 x °+ 12 

27. r 2 — rs — 90 s 2 

39. 

w 2a — 10 w a + 24 

28. m 4 — 10 m 2 p — 24 p 2 

40. 

a 2v - 11 a y + 30 

29. x 4 — 2 x 2 y — 63 y 2 

41. 

c 2z -12 c* + 35 

30. t 2 + 2 tw - 48 w 2 

42. 

x 2n — 2 — 35 

31. a 4 — a 2 b — 20 b 2 

43. 

2/26 4- 4 ^ - 21 

32. c 6 + c 3 — 56 

44. 

z 2a — 3 z a — 54 


45. Factor c 2 — A c — .21. 

Solution. 1. Two numbers whose product = — .21 and whose 
sum = — .4, are — .7 and + .3. 

2. /. c 2 - .4 c - .21 = (c - .7)(c + .3). 


46. m 2 + .9 m + .20 

47. n 2 - .8 n +.12 

48. r 2 + .5 r — .14 

49. p 2 — .6 p — .27 


60. (x + y) 2 — b(x + y) + 6 

61. (c - d ) 2 + 7(c - d) + 12 

62. ( a — b) 2 + 2(a — 6) — 3 

63. (r + s) 2 - 3 (r + s) - 10 


122 


ALGEBRA 


98.* Factoring trinomials of the form mx 2 + nx + p- 

The C.E.E.B. does not, apparently, require this in Elementary Alge¬ 
bra. In many respects, it is the most important case of factoring. 

The product of two binomials like (2 x + 3) and (3 x + 5) is a 
binomial of the form mx 2 + nx + p. 

This means that there is a term containing the second power 
of x, one containing the first power (as a rule), and one free 
from x. The following discussion shows how to factor tri¬ 
nomials of the form mx 2 + nx + p, when they are factorable. 
Development. Factor 12 x 2 + 23 x 4* 5. 

Solution. 1. The first terms of the binomials may be 2 x and 6 x, 
for their product is 12 x 2 . Place them in parentheses thus : 

(2 x )(6x ). 

The second terms must be 1 and 5, since these are the only factors 
of 5. 

2. The second terms of the binomials are both positive since 5 and 
23 x are positive. Place the factors + 5 and + 1 in the parentheses 
and note the middle term which results. 

a. (2 x + 5) (6 x + 1); middle term, + 32 x. Incorrect. 

b. At once, interchange 5 and 1 in the parentheses. Thus : 

(2 x + 1) (6 x + 5); middle term, + 16 x. Incorrect. 

3. Step 2 shows that the factors 2 x and 6 x for 12 x 2 are incorrect,. 

Try 3 x and 4 x for 12 x 2 , thus : (3 x )(4x ). 

a. (3 x + 1) (4 x + 6); middle term, + 19 x. Incorrect. 

b. (3 x + 6) (4 x + 1); middle term, + 23 x. Correct. 

Check. Does (3 x + 5)(4 x + 1) = 12 x 2 + 23 a: + 5? Yes. 

EXERCISE 70* 

Factor: 


1. 

2 

X 2 

+ 

7 x T 3 

6. 

4z 2 

+ 8 x + 3 

2. 

6 

X 2 

+ 

11 x + 3 

7. 

6 y 2 

+ 11 y + 3 

3. 

6 

r 2 

+ 

5 r + 1 

8. 

2x 2 

+ 9x + 10 

4. 

2 

s 2 

+ 

7s + 5 

9. 

3 z 2 

+ 10 2 -f- 8 

6. 

3 

a 2 

+ 

8 a + 5 

10. 

12 v? + 17 w + 6 


Solution. Since 2 is positive, its factors must have like signs 
and, since — 7 is negative, the cross products must both be negative. 
Therefore both factors of 2 must be negative. (Complete the solution.) 


SPECIAL PRODUCTS AND FACTORING 123 


11 . 5 m 2 — 7 m + 2 

12 . 3 m 2 — 7 m + 4 

13. 4 p 2 — 9 p + 5 

14. 9 s 2 —9 5 + 2 

15. Qt 2 - 1U + 4 


17. 3 w 2 - 17w + 20 

18. 6 y 2 - 19 y + 3 

19. 14 y 2 - 11 y + 2 

20 . a ? 2 — 10 xy + 24 y 2 


21 . 9 x 2 — 12 xy + 4 y 2 


16. 2 w 2 - 11 w + 15 


22 . Factor 15 a : 2 + 14 x — 8 . 


Solution. 1. Since — 8 is negative, its factors must have unlike signs; 
since 14 x is positive, the larger cross product must be positive. 

2. For 15 x 2 , try (3 x )(5x ). For 8, try 2 and 4. 

a. (3 a; 2)(5x 4). The cross product, 3 x • 4 is the larger; 

therefore make 4 positive and 2 negative. 

Thus: (3 x — 2) (5 x + 4). Middle term, 2 x. Incorrect. 

b. At once, try (3 x 4) (5 x 2). Now the cross product, 
4 • 5 x, is the larger. Make 4 positive and 2 negative. 

Thus: (3 x + 4) (5 x — 2). Middle term, + 14 x. Correct. 

23. Factor 24 m 2 — m — 10. 

Solution. 1. The factors of — 10 must have unlike signs. The 
signs must be selected so that the larger cross product is negative, 
since — m is negative. 

2. For 24 m 2 , try (6 m )(4 m ). For 10, try 5 and 2. 

a. (6 m — 5) (4 m + 2). Middle term, —8 m. Incorrect. 

b. At once, try interchanging 5 and 2. 

Thus: (6m + 2)(4m -5). Middle term, — 22 m. Incorrect. 

3. For 24 m 2 , try 3 m and 8 m. For 10, try 5 and 2. 

a. Thus : (3 m — 5) (8 m + 2). Middle term, — 34 m. Incorrect. 

b. Try (3 m — 2) (8 m + 5). Middle term, — m. Correct. 

Note. — First select the proper factors to give the end terms; then 
decide which cross product is the larger; then give the second terms their 
signs so that this larger cross product has the same sign as the middle term. 


24. 2 x 2 + 5 x — 3 

25. 6 a 2 — 5 a — 6 

26. 3 a 2 + 8 a — 3 

27. 4 b 2 — 5 b — 6 

28. 5 c 2 — c — 4 


29. 6 d 2 + 7 d — 5 

30. 5 k 2 + 7 h — 6 

31. 12 m 2 — m — 6 


32. 12 p 2 + 8 p — 15 

33. 12 x 2 — 13 xy + 3 y 2 


(Continued on page 124.) 


124 


ALGEBRA 


34. 

18 x 2 

+ 3 X2 - 10 Z 2 

48. 

121 - 

- 44 x + 4 x 2 

35. 

8 t 2 - 

-26 tw + 21 w 2 

49. 

x 2 + 

xy — 56 y 2 

36. 

16 KYV 

! 4- 18 TOM —9 m 2 

50. 

18 x 2 

- 39 x +11 

37. 

x 2 - 

5 xy — 24 j/ 2 

51. 

14 r 2 

- 61 rt - 9 t 2 

38. 

X 2 — 

wx — 110 w 2 

52. 

63 s 6 

+ s 3 - 20 

39. 

12 r 2 

- 13 rs - 14 s 2 

53. 

40 y 2 

+ yw — 6 w 2 

40. 

9 a 2 - 

- 30 ab + 25 V 

54. 

21 - 

40 x + 16 x 2 

41. 

63 + 

2 x — x 2 

55. 

27 o 2 

-f- 15 ab — 50 b 2 

42. 

14 - 

29 y — 15 y 2 

56. 

40 x 2 

- 18 x - 9 

43. 

16 - 

40 x + 25 x 2 

57. 

36 to 2 

; — 84 mn + 49 n‘ 

44. 

8 x 2 - 

- 18 x 1 / + 7/ 

58. 

12 r 2 

-43 rt - 20 t 2 

45. 

10 z 4 

- 21 2 2 + 9 

59. 

20 m 2 

: + 9 mn — 20 n 2 

46. 

24 c 2 

— cd — 10 cP 

60. 

21 A 1 

! + 2 A - 8 

47. 

28 r 2 

+ 3 rs - 18 s 2 

61. 

4 x 2 - 

- 24 x + 35 


Case IV 

Type Form: (a + b) 2 = a 2 + 2 ab + b 2 

Note. — This section and the next are required as preparation for 
§ 101, p. 127; and the latter is required by the College Entrance 
Examination Board, and is probably implied by the Syllabus issued by 
the Regents in 1928. 

99. Squaring a binomial. 

Development. 1. (a + b) 2 = (a + b)(a + b) 

= a 2 + 2 ab + b 2 . (§ 96) 

2. Similarly, ( a — b) 2 ={a — b)(a — b) 

= a 2 - 2 ab+ b 2 . 

Rule 1 . — To square the sum of two numbers : 

Square the first number; add twice the product of the first 
and second numbers; add the square of the second number. 

Rule 2 . — To square the difference of two numbers : 

Square the first number; subtract twice the product of the 
two numbers; add the square of the second number. 


SPECIAL PRODUCTS AND FACTORING 125 


Example 1. Find the square of 3 x 2 — 4 y z . 
Solution. 1. Using Rule 2, above: 

2. (3 x 2 - 4 y*) 2 = (3 x 2 ) 2 - 2(3 x 2 )(4 y*) + (4 y*) 2 

3. = 9 x 4 - 24 xV + 16 2/ 6 . 

Example 2. Find the square of .3 x + .2 y. 
Solution. 1. Using Rule 1, above: 

2. (.3 x + .2 </) 2 = (.3 x) 2 + 2(.3 x)(.2 y) + (.2 y ) 2 

3. = .09 x 2 + .12 xy + .04 y 2 . 

Note. — Step 2 of each solution should be done mentally. 

EXERCISE 71 

Square the following binomials by inspection: 


1. a + 7 

11. 

4 a x — b 2 

21. Qt + 9 n 


2. b -|- 8 

12. 

3 c v + d 2 

22. .3 - 4 x 


3. c — 5 

13. 

4 x° - 3 y b 

23. 5 - .7 x 


4. d - 4 

14. 

5 x a + 2 y b 

24. 8 y — .3 x 


6. <? + 9 

15. 

3 x 2 - 5 y 2 

25. 3 cd — 10 


6. x 2 + 10 

16. 

2 x 2 - 7 y 3 

26. .5 r 2 — 9 5 


7. r 2 - 11 

17. 

3 x 2 + 6 d 

27. .2 x 2 — .6 y 2 


8. s 2 - 12 

18. 

5 m + 8 n 3 

28. 7 c 3 — 4 dm 


9. ab + 20 

19. 

3 xy — 7 

29. 11 a 2 - .3 b z 


10. 2 a 2 + 3 

20. 

4w 2 — 5 z 

30. 12 m 2x -5 n v 


31. m + \ 

33. x - 

~i 35. 2 - 

- | 37. a - f 


32. n + i 

34. y + f 36. w 2 

+ f 38. b + | 



39. Square 29 mentally. 


Suggestion. 

29 2 = (30 - l) 2 = 900 - 

-60 + 1, or 841. 


Similarly square : 




40. 12 

43. 49 

46. 38 

49. 22 52. 

43 

41. 19 

44. 18 

47. 21 

50. 42 63. 

62 

42. 39 

45. 28 

48. 31 

51. 23 54. 

81 


126 


ALGEBRA 


100. A perfect square trinomial is the result obtained from 
squaring a binomial. 

Thus, (3 * - 5 y) 2 = 9 x 2 - 30 xy + 25 y\ 

Then 9 x 2 — 30 xy + 25 y 2 is a perfect square trinomial. 

Rule. — 1. A trinomial is a perfect square when two of its 
terms are perfect squares (9 x 2 and 25 y 2 ) preceded by plus 
signs; and when the remaining term (30 xy) is twice the prod¬ 
uct of the square roots of the perfect square terms, preceded by 
either a plus sign or a minus sign. 

2 . To find the square root of a perfect square trinomial: 
extract the square roots of the two perfect square terms, and 
connect them by the sign of the remaining term. 

Example. Express 4 x 2 + 9 y 4 — 12 xy 2 as the square of a 
binomial. 

Solution. 1. V4 x 2 = 2 x\ V + 9 y* — 3 y 2 ; 12 xy 2 = 2(2 x) • (3 y 2 ). 

2. /. 4 x 2 + 9 ?/ 4 — 12 xy 2 = (2 x - 3 y 2 ) 2 . 

EXERCISE 72 

Express each of the perfect squares of the following list as the 
square of a binomial. 


1. 

X 2 

+ 6z + 9 

15. 

X 2 

4“ 8 x -f- 16 

2. 

m 

2 - 10 m + 25 

16. 

4 X 2 + 12 X + 9 

3. 

y 2 

- 4 y + 4 

17. 

4 a 2 - 20 ax + 25 x 2 

4. 

X 2 

+ 12 x + 36 

18. 

9 a 

4 - 30 xy + 25 y 2 

5. 

4. 

7—H 

+ 

a 

+ 

ca 

& 

19. 

25 

y 2 + 10 yz + z 2 

6. 

9; 

x 2 — 6 x + 1 

20. 

9 m 2 + 42 mn + 49 n 2 

7. 

z 2 

- 16 z + 64 

21. 

49 

y 2 — 14 yz + z 2 

8. 

36 

i + 12 s + s 2 

22. 

m 2 

— 24 mn + 144 n 2 

9. 

c 2 

— 6 cd + 9 d 2 

23. 

x« 

— 4 xhy + 4 y 2 

10. 

X 2 

+ 4 xy + 4 y 2 

24. 

x 2a 

— 6 x a + 9 

11. 

1 

— 6 a + 9 a 2 

25. 

y20 

- 8 y c + 16 

12. 

c 2 

d? — 2 cd 

26. 

121 x 2 - 44 xy + 4 y 2 

13. 

X 2 

+ 25 y 2 - 10 xy 

27. 

81 

a? + 4 b 2 - 36 ab 

14. 

36 

a 2 — 12 ab + b 2 

28. 

z 2n 

+ 10 z" + 25 



SPECIAL PRODUCTS AND FACTORING 127 


101. Polynomials reducible to the difference of two squares. 

Example t. Factor a 2 — 2 ab + b 2 — c 2 . 

Solution. 1. a 2 — 2 ab + b 2 — c 2 = (a — b) 2 — c 2 
2. = {a — b c){a — b — c). 

Example 2. Factor a 2 — b 2 + 2 be — c 2 . 

Solution. 1. a 2 — b 2 + 2 be — c 2 = a 2 — (5 2 — 2 be + c 2 ) 

2. = a 2 — (6 — c) 2 

3. = [a +(fr — c)][a — (6 — c)] 

4. = (a + & — c) (a — & + c ). 


Factor: 

1. (x + y ) 2 — z 2 

2 . (2 a - 6) 2 - 1 

3. (a + 5) 2 -4c 2 

4. (2 x 2 — 3 7/) 2 — 9 

5. (c + 3) 2 -4 6ft 2 

6. (a + &) 2 — c2 

7. (m — w) 2 — 4 p 2 

8. (z+3) 2 -2/ 2 

9. (a - 3 &) 2 - c 4 

10. 16 a 2 + b 2 + 8 — c 2 

11 . 25 £ 2 — 4 £ 2 10 £ + 1 

12. 4 z 2 + 4 x — y 2 + 1 

13. a 2 b 2 + 8 ab + 16 - g 2 

14. 9 x 2 + 6 x — 16 y 2 + 1 

15. M 2 - K 2 + N 2 + 2 MN 

16. m 2 + 2 mn + n 2 — 81 

17. d 2 + 1.4 d + .49 - w 2 

18. 4 y 2 + 20 y + 25 — .25 x 2 

19. **» + 2 a* + 1 2/* 

20. 6 4 — 6 6 2 c + 9 c 2 — 4/ 2 


s 73 

21. p 2 — 4 pq + 4 g 2 — 25 

22. £ 2 — (?/ + z ) 2 

23. x 2 — (y — z ) 2 

24. 4 a 2 — (b — 2 c) 2 

25. 9 — (a; + a) 2 

26. 16 m 2 - (n + l) 2 

27. 1 — (x — y) 2 

28. t 2 — r 2 — 2 rs — s 2 

29. 4 c 2 — a 2 + 2 ab — b 2 

30. 9 r 2 — 4 s — 4 — 4 s 2 

31. 25 / 2 — 4 z 2 + 6 xy — 9 y 2 

32. 8 * + 49 y 2 - x 2 - 16 

33. 36 — 12 ab — a 2 — 36 b 2 

34. 81 z 2 + 20 w — 4 w 2 — 25 

35. m 2 — 4 r 2 — 8 rx — 4 z 2 

36. c 2 — b 2 — 4 a 2 — 4 ab 

37. 4 + 22 xy — 121 z 2 — y 2 

38. 4 a 4 — x 2 — 9 y 2 + 6 xy 

39. 2 AB - A 2 + T 2 - B 2 

40. 1 — & 2 — 9 + 6 b 


128 


ALGEBRA 


EXERCISE 74# 

Factoring Polynomials by Grouping 

Note. — This topic is partially required by the College Entrance 
Examination Board, namely to the extent of examples like numbers 
3 to 20 below. It is not explicitly mentioned in the Syllabus issued by 
the Regents in 1928, but it is probably implied. 

Example 1. Just as ax + bx = (a + 6)x 
so a(x + y)+ b(x + y) = (a + b)(x + y). 

Example 2. Factor 6 x 3 — 15 x 2 — 8 x + 20. 


Solution. 1. 6 x 3 — 15 x 2 — 8 x + 
2. 

3. 

Factor: 

3. 2 a(x + y)- 30 + y) 

4. 5 m(r + s)+ 2 n(r + s) 

5. 3 p(2 x — y)— r(2 x — y) 

6. 8(t + w)— m(t + w) 

7. a(6 + c) — d(b + c) 

8. a(6 + n)-\- m(b -(- n) 

9. a(x - y)+ b{x - y) 

10. a(c — d)— b(c — d) 

11. a?(a + l)+(o + 1) 

12. x 2 (4 x — 5)—(4x— 5) 

13. x 2 (x +1)— 20 + 1) 

14. y(y2 + 3)-(y 2 + 3) 

15. z 2 (z - 2)- 2(2 - z) 

16. m(m 2 - 3)- 2(3 - m 2 ) 

17. c(c 2 +5)-3(5 +c 2 ) 

18. xy(x — 2)+ 40 — 2) 

19. 2 (cd - 5)- c{cd - 5) 

20. c(a + 26)— d(a + 26) 


20 = (6 x 3 - 8x)-(15x* - 20) 

= 2 x(3 x 2 - 4) - 5(3 x 2 - 4) 
= (2x - 5) (3 x 2 - 4) 

21. ac + ad — 3 bd — 3 be 

22. mr — 2 nr + ms — 2 ns 

23. 3 ax — 6 ay — bx + 2 by 

24. 2mr-3ms — 3ns-\-2nr 

25. 2 + 3 a — 8 a 2 — 12 a 3 

26. 3a; 3 +6;r 2 +£ + 2 

27. 10 mx— 15 nx— 2 ra+ 3 n 

28. a?x-\~abcx—a 2 by—b 2 cy 

29. a 2 bc—ac 2 d-\~ab 2 d—bcd? 

30. 30 a 3 - 12 a 2 - 55 a + 22 

31. 56 — 32 x + 21 x 2 — 12 x 3 

32. 3 ax — ay + 9 bx — 3 by 

33. 4 x 3 + x 2 y 2 — 4 y 3 — 16 xy 

34. rt — m — sn + st 

35. 2 cx — 3 cy + 6 dy—4 dx 

36. x 3 — xy 2 + y 3 — x 2 y 

37. m 3 — n 3 — m 2 n + mn 2 

38. ax + 2 ay — 3 6x — 6 by 


SPECIAL PRODUCTS AND FACTORING 129 

EXERCISE 75 (Review of Factoring) 

Find all the prime factors of the following: 

29.* z 2 + 2 zt - 63 t 2 


1 . 

2. 9 x 2 y — 16 y 

3. ax 2 — ay 2 + az 2 

4. 5 x 2 - f y 2 

5. (a — 2 b) 2 — c 2 

6. ax 2 — 2 ax — 5 a 

7. 9 r 2 s - 16 s 3 

8. a(x - y)~ b(x - y) 

9 1 n 4: _ 9 «2 

* 2 5 a 16 C 

10. mx 3 — my 3 + ms 3 

11. x 2 -(2 x- l) 2 

12. 20 m 2 r — 125 n 2 r 

13. 2 mx — 2 my + 2 mz 

14. .3 x — .3 y + .3 z 

15. 9a— am 2 

16. a 2x - b 4 ” 

17. 5 pr 2 + 5 pr — 5 p 

18. 4 a 1 - 9 a x b 2 

19. .5 x — 1.5 y + 2.5 z 

20. 4 y 2 — {x + z) 2 

21. 9 x a m 2 — x a 

22. c(m + n)— 2 d(m + n) 

23. x 2n — {y + z) 2 

24. 7 x 5 — 7 x 

25. .5 r 4 — .5 z 4 

26. ax 2 — a (a; — l) 2 

27. x 2n — 4 t/ 2 

28. 18 be 2 — 2 for 2 


30. * y 2 - 8 y + 12 

31. * a 2 + 5 ab — 36 6 2 

32. * && 2 + 9 cd - 52 

33. * m 2 — 7 mo: — 44 £ 2 

34. * a; 2 + 2 a??/ + i/ 2 

35. * 15 b 2 + b - 16 

36. * * 2 + !•* + * 

37. * w 2 + 72 - 18 w 

38. a 2 — 2 ab + b 2 — (? 

39. * fo 2 - 7 & + 12 

40. * c 2 - 19 c + 18 

41. * z 2 + 100 - 25 x 

42. * ac — be ad — bd 

43. z 2 — y 2 — 4 y — 4 

44. * 6 r 2 + 7 r - 5 

45. * 2 aw 2 — 3 aw — 20 a 

46. * az — 2a— bx -{- 2 b 

47. ir 2 — 6 £ + 9 — y 2 

48. * 42 t - 15 tx - 3 to 2 

49. * a 2 — 50 a + 49 

50. * w 2 — 26 w + 169 

51. * 6 mx 2 — 13 mxy + 6 my 2 

52. * 4 x 2 y — 12 xy 2 + 9 y 3 

53. * ax — 2 ay + 3 bx —6 by 

54. * 4 c 2 d — 8 cd — 21 d 

55. 4 x 2 — 4 xy + y 2 — 9 z 2 
66.* a 2 c — 36 b 2 c + 5 abc 


Note. — Exercises 161-163, pages 283 to 285, can be done now. 


130 


ALGEBRA 


102. Simplification by use of special products. Do all the 
multiplying mentally; write only the results. 

Example. Find the simplest form of 

3 a(2 a + 7)(a — 3) — 4 a(a + 5)(a — 6). 

Solution. 1. 3 a(2 a + 7) (a — 3) — 4 a(a + 5) (a — 6) 

2. =3 a(2 a 2 + a — 21) — 4 a{a 2 — a — 30) 

3. = 6 a 3 + 3 a 2 - 63 a - 4 a 3 + 4 a 2 + 120 a 

4. = 2 a 3 + 7 a 2 + 57 a. 

Note 1. — The equality sign in line 2 means that the expression in 
line 1 equals that in line 2, etc. 

Note 2. — The binomials are multiplied together in Step 2, and then 
their products are multiplied by the monomials in Step 3. 


EXERCISE 76 

Find (without use of any scratch paper): 


1. 2 m(4 ra — l)(ra + 2) 

2. — 3(5 — z)(6 + x) 

3. + 4(2 c — 3 d)(2 c — 3 d) 

4. - 5{x - 4y)(x +4 y) 


5. + 6(3 a — 2 b) (a + 3 b) 

6. — 2(2 r +5s)(2r -5 s) 

7. + 8(m — 11 n){m + 4 n) 

8. — a(8 a -7m)(2a + 3m) 


Solve the following equations and problems: 

9. 2(3 x - l)(a + 2) - 3(a - 2)(2 x + 1) = 40 
Solution. 1. 2(3 x — \){x + 2) — S(x — 2)(2 x + 1) = 40. 

2. .*. 2(3x 2 + 5x -2) - 3(2 rr 2 - 3 x - 2) = 40. 

3. /. 6z 2 + 10a: —4-6x 2 + 9z + 6 = 40. 
(Complete the solution.) 

10. (4 y - Z){y + 2) - 4 (y - 3)(t/ + 3) = 35 

11. (2-3 0(2 - 3 t) - (3 t - 1) (3 t + 1) = 2 

12. 5(2 r - 1)(3 r + 1) - (5 r + 2)(6 r - 3) = 2 

13. 3(2 a - 5)(o + 4) - 6(o + 8 )(a - 2) = 9 

14. 7(a - 1)(2 a + 1) - (7 a - 1)(2 a + 1)-2 

15. (5 x — 4)(5 x - 4) - 5(5 x + 2)(x - 4) = 6 

16. (5 x - 4)(3 x + 4) - 3(5 x + 6)(z - 7) = 10(9 x + 15) 


SPECIAL PRODUCTS AND FACTORING 131 

17. (3m + l)(m - 3) - 3(m - 4) (m + 2) = 25 

18. 2(5 + 5) (3 5 - 4) - 6(5 - 6) 2 = 26 

19. (z - 4) (z + 4) - z(z + 6) = 2 (z + 8) 

20. 6 w (2 w + 1) - (4 w - 5) (2 w + 3) = 2 w (2 w — 13) 

21. A certain number exceeds another number by 5. The 
square of the larger exceeds the square of the smaller by 95. 
What are the numbers ? 

22. Separate 25 into two parts such that the square of the 
larger exceeds the square of the smaller by 225. 

23. A certain number is 8 more than another number. The 
product of the two numbers exceeds the square of the smaller 
by 40. What are the two numbers ? 

24. There are three consecutive integers such that the product 
of the second and third exceeds the square of the smallest by 47. 
What are the integers ? 

25. There are three consecutive integers such that the square 
of the second, diminished by the square of the first, gives a 
remainder which is 11 less than the third. What are the integers ? 

26. Find three consecutive integers such that twice the 
square of the third exceeds the sum of the squares of the 
first and second by 43. 

27. (3 s- 5) (2 x + 4) = (3 x — 4) (2 x + 7) - 3 

28. (6 a + 1)(3 a - 4) - 2(3 a - 2) 2 = 3 

29. (2 m — 5) 2 = (4 m — 3)(m + 1) 

30. (5 - 3 5)(2 + s) - 3(1 - s)(l + s) = 10 

31. 2(5 t - 2)(3 t + 1) - 3(2 t + 3)(5 t - 2) = 49 

32. 3 y(y + 4) - (y + 1)(2 y - 1) = (y + 1 )(y + 9) 

33. 5(2 a - l) 2 - (4 a - 3)(4 a + 3) = 2(2 a + 7)(a - 5) 

34. (1 - 3 xf - (2 x - l) 2 = 5(x + 1) (re - 7) - 21 

35. Separate 15.5 into two parts such that the square of the 
larger exceeds the square of the smaller by 69.75. 


132 


ALGEBRA 


EXERCISE 77 

Problems about Areas 

1. a. What is the area of the rectangle whose base is (x — 2) 
inches and altitude is (x + 5) inches ? 

b. What is the perimeter of this rectangle ? 

2. The base of a certain rectangle exceeds its altitude by 5 
inches. 

a. If the altitude is h in., what is the length of the base ? 

b. What is the area of the rectangle ? 

c. What is the area of the square whose side equals the base 
of this rectangle ? 

d. Write the equation which expresses the fact that the area 
of the square is 65 sq. in. more than the area of this rectangle. 

e. Solve this equation and find the dimensions of the rectangle 
and of the square. 

Note. — The essential facts of this example can be charted thus: 



Base 

Altitude 

Area 

Rectangle 

h 4-5 

h 

h(h 4~ 5) 

Square 

h 4-5 

h 4~ 5 

(h 4- 5) 2 


.*. (h 5) 2 — h(h 4~ 5) -j- 65. 


3. a. Express the area of the square whose side is s inches. 

b. Express the base, altitude, and area of the rectangle whose 
base is 10 inches longer and whose altitude is 2 inches longer 
than the side of the square. 

c. Arrange the results of parts a and b in a chart as in the 
Note above. 

d. Write the equation which expresses the fact that the area 
of the rectangle is 200 sq. in. more than the area of the given 
square. 

e. Solve the equation, and find the dimensions of the square 
and the rectangle. Check the solution. 







SPECIAL PRODUCTS AND FACTORING 133 


4. The base of a rectangle exceeds its altitude by 9 inches. 
A better shaped rectangle is formed by making the altitude 3 in. 
longer and the base 2 in. longer; but this increases the area by 
78 sq. in. What are the dimensions of the two rectangles? 
(Represent the dimensions; form an equation, and solve it.) 

5. The base of a rectangle exceeds its altitude by 8 inches. 
If the base and altitude are both decreased by 4 inches, the 
old area exceeds the new area by 192 square inches. What are 
the dimensions of the rectangle ? 

6. The base of a rectangle is 7 feet more than its altitude. 
If the base be increased 5 feet and the altitude be decreased 
3 feet, the area will remain unchanged. What are the dimen¬ 
sions of the rectangle ? 

7. The base of a rectangle exceeds twice its altitude by 3 
inches. The area of a square whose side equals the base exceeds 
twice the area of the rectangle by 57 square inches. What are 
the dimensions of the rectangle? 

8. A man had plans for a house whose foundation was 12 
feet longer than it was wide. Finding that it was too expensive, 
he cut off 3 feet from the length and 2 feet from the width. The 
new foundation covered 158 square feet less than the old. What 
were the dimensions before and after making the change ? 

9. The main shaft of Washington Monument is square at 
the bottom and top. The side of the lower square exceeds the 
side of the upper square by 21 feet. The area of the lower 
square exceeds the area of the upper square by 1869 square feet. 
Find the dimensions of the two squares. 

10. The base of a rectangle exceeds its altitude by 7 inches. 
A second rectangle is formed by making the altitude 3 inches 
shorter and the base 10 inches longer. The new rectangle 
contains 75 square inches more than the first rectangle. What 
are the dimensions of the first rectangle ? 


134 


ALGEBRA 


103. * Quadratic equations solved by factoring. 

Note. —An optional topic in the Regents’ 1928 Syllabus. 

Example. Find the number whose square exceeds the num¬ 
ber itself by 6. 

Solution. 1. Let n ='the number. 

2. Then n 2 = the square of the number. 

3. Then n 2 = n + 6. 

4. .*. n 2 — n — 6 = 0. 

5. Factoring, (n — 3) (n + 2) = 0. 

6. If n has the value which makes n — 3=0, 

then (n — 3)(n + 2) = 0, for 0 • (n + 2) = 0. 

But n — 3=0, if n = 3 

Also, if n has the value which makes n + 2=0, 
then (n — 3) (n + 2) = 0, for (n — 3) • 0 = 0. 

But n + 2=0, if n = — 2. 

It seems, then, that + 3 and — 2 are both roots of the equation. 

Check. Does 3 2 - 3 - 6 = 0? Does 9-9=0? Yes. 

Does (- 2) 2 -(- 2)- 6 = 0? Does 4 + 2 - 6 = 0? Yes. 

An equation like n 2 — n — 6 = 0 is called a quadratic equa¬ 
tion or an equation of the second degree. 

Other examples are: 4 x 2 — 9 = 0 

and 3y 2 = \y + 9. 

5 6 

Notice that the equation has only one unknown; that this unknown 
does not appear in the denominator of any fraction; that it does appear 
with exponent 2; that it may or may not appear with exponent 1. 

Every quadratic equation has two roots, just as the equation 
n 2 — n — 6 = 0. 

104. * Solution of equations by factoring depends upon the 
following fact: 

If one of the factors of a product is zero, the product also is zero 
Thus,3 X 0 = 0; (- 5)X 0 = 0; 2X0 X(- 3) = 0. 

Moreover, if a product of two or more factors is zero, then 
one, or more, of the factors must be zero. 

Thus, if (x — l)(x + 2) = 0, then x — 1 must = 0, or x + 2 must = 0. 


SPECIAL PRODUCTS AND FACTORING 


135 


105.* Rule. — To solve an equation by factoring : 

1. Transpose all terms to the left member. 

2. Factor the left member completely. 

3. Set each factor equal to zero, and solve the resulting equa¬ 
tions. 


4. The roots obtained in Step 3 are the roots of the given 
equation. 

5. Check by substitution in the given equation. 


Example. Solve the equation x 2 = 20 — 8 x. 


Solution. 1. 

2. Transpose 

3. Factor 

4. 

5. 


x 2 = 20, - 8 x 
x 2 + 8 x - 20 = 0 
(x + 10) (x - 2) = 0 
x + 10 = 0, if x = — 10 
x — 2 = 0, if x — 2. 

.*. x = — 10; or x = 2. 


Note. — Step 4 should be done mentally, and omitted. 


EXERCISE 78 * 


Solve by factoring the following quadratic equations: 

1. z 2 - 10 x + 21 = 0 15. 4 c 2 + 5 c - 6 = 0 

16. 5 m 2 — 6 = 13 m 


2. 

X 2 

+ 

7 x + 10 = 

3. 

y 2 

- 

5 y — 24 = 

4. 

t 2 ■ 


50 

1 

00 

5. 

s 2 

+ 

75 = 44 

6. 

X 2 

— 

x = 42 

7. 

y 2 

= 

18 y — 80 

8. 

w 2 

— 

36 = 5 w 

9. 

X 2 

— 

25 = 0 

10. 

TO 2 


■ 49 = 0 

11. 

z 2 

= 

5 z + 14 

12. 

X 2 

= 

a 

1 

o 

05 

13. 

yl 

- 

-4 

<si 

II 

t—i 

00 

14. 

X 2 

— 

5 x = 66 


17. 

2 

a 2 

+ 

7a - 4 = 

0 

18. 

3 

X 2 

— 

7 x -f- 2 = 

0 

19. 

2 

X 2 

= 

5 x + 3 


20. 

T 

l _ 

- 9 

x = 0 


21. 

2 

y 2 

+ 

o 

II 

>5 


22. 

5 

TO 2 


■ 27 m = 18 

23. 

7 

X 2 

- 

TJH 

II 

a 

CO 


24. 

5 

y 2 

+ 

3 2/ — 2 = 

0 

25. 

8 

z 2 

= 

5 z + 3 


26. 

3 

t 2 


II 

T—1 


27. 

8 

X 2 

= 

1 - 2x 


28. 

3 

a 2 

+ 

13 a = 10 



136 


ALGEBRA 


106.* Problems solved by quadratic equations. 

Example. Find three consecutive integers such that twice 
the square of the second, increased by the product of the first 
and the third, is 296. 

Solution. 1. Let x = the first of these three integers. 

x + 2 = the second of these three integers 
and x + 4 = the third of these three integers. 

2. .'. 2(x + 2) 2 + x(x + 4) = 296. 

3. 2 z 2 + 8 x + 8 + z 2 + 4 x - 296 = 0. 

4. .'. 3 z 2 + 12 x — 288 = 0. 

5. D 3 z 2 + 4 x — 96 = 0. 

6. Factoring, (x + 12) (x — 8) = 0. 

7. .'. x = — 12, or x = 8, the first of the three integers. 

8. When the first integer is — 12, the next two even integers are 
— 10 and — 8. 

Check. Does 2(- 10) 2 +(- 12)(- 8) = 296? 

Does 200 + 96 = 296? Yes. 

9. When the first integer is 8, the next two even integers are 10 
and 12. 

Check. Does 2(10) 2 + 8 X 12 = 296? 

Does 200 + 96 = 296? Yes. 

In this example, two sets of even integers are obtained, which satisfy 
the conditions of the problem. 

EXERCISE 79* 

1. The square of a number equals the sum of 5 times the 
given number and 6. What is the number ? 

2. The square of a number diminished by 8 times the number 
itself is 33. What is the number ? 

3. If the square of a number be increased by 9 times the 
number and 25, the result is 7. What is the number ? 

4. The square of a certain number increased by the square 
of the number which is 2 more than the first gives the result 100. 
What are the numbers ? 

5. John is 4 times as old as Charles. If John’s age be added 
to the square of Charles’ age, the sum is 45. What are their 
ages? 


SPECIAL PRODUCTS AND FACTORING 137 


6. The sum of the squares of two consecutive integers is 145. 
What are the integers ? 

7. There are three consecutive integers such that the square 
of the first, increased by the square of the second, and this result 
diminished by the square of the third, gives 96. What are the 
integers ? 

8. There are three consecutive odd integers such that the 
sum of the squares of the second and third, increased by twice 
the square of the first, equals 300. What are the integers ? 

9. Separate 22 into two parts such that the square of the 
larger increased by the square of the smaller is 260. 

10. Twice the square of a number equals the number itself. 
What is the number ? 

11. 9 times the square of a certain number is 25. What is 
the number ? 

12. The area of a certain small piece of ground is 240 square 
rods. Its length is 8 rods more than its width. What are its 
dimensions ? 

13. How long and wide must a lot be to contain 4500 square 
feet, and have the length 5 times the width ? 

14. What are the dimensions of a triangle whose area is 54 
square inches if the base is 3 times as long as the altitude ? 

15. A certain lot is 30 feet wide and 140 feet long. By in¬ 
creasing the width by a certain number of feet, and decreasing 
the length by twice the same number of feet, the area is increased 
350 square feet. What is the number of feet ? 

16. What must be the dimensions of a lot containing 8000 
square feet, such that the length is 10 feet more than 3 times 
the width ? 

17. There are three consecutive integers such that the square 
of the first, increased by the square of the second, is 115 less 
than twice the square of the third. What are the integers ? 


138 


ALGEBRA 


REVIEW EXERCISE VIII 

1. From + 5 x 3 — 3 x 2 + 4 subtract 2 x 3 — 3 x 2 — 5 x + 8, 
and find the numerical value of the remainder when x = 2. 

2. Simplify 2a— 3 (a — 1) — [a — (2 a — 1)]. 

3. Divide 6 x 3 + 5 x 2 — 4 x + 2 by 2 x + 3, and write the 
result as a mixed expression. 

4. Is — 2 a root of the equation 3x + 5 = x+ l? 

5. a. If m apples cost x cents, what will n apples cost? 

b. A is five times as old as B. If x equals B’s age now, how 

old will A be in 4 years ? How old was B six years ago ? 

6. a. By the formula S = - • (2 t — 1). Find t when a = 14 

and S = 147. 2 

b. If t has a fixed value, how does S change when a is 
doubled ? 

7. a. Express by algebraic symbols: the square of a dimin¬ 
ished by the cube of b. 

b. What is the total cost of m pencils at x cents each and 
n books at y cents each ? 

8. The sum of two numbers is 110. If five times the smaller 
be added to three times the larger, the sum is 390. Find the 
numbers. 

9. A certain man’s age 10 years from now will be five times 
what his age was 10 years ago. What is his age now ? 

10. A and B are traveling in opposite directions at the rate of 
22 miles and 27 miles per hour, respectively. If they started 
from the same place at the same time, in how many hours will 
they be 147 miles apart ? 

11. Which of these statements are true and which false? 

d. = 1, if x — 3. 
x 

e. \ gt 2 = 64, if g = 32 and t = 2. 
/. xy = 6, if x = 6 and y = 9. 


a. 3 x = 12, if x = 4. 

b. x 3 = 12, if x = 4. 

c. 3 + x = 12, if x = 4. 


IX. FRACTIONS 


107. The expression ^ is a fraction ; it is read “ a divided 

by b” a is the numerator, and b is the denominator of the 
fraction; together, they are called the terms of the fraction. 
The denominator, b, must never be zero. (See § 67) 


REDUCTION OF FRACTIONS TO LOWEST TERMS 

108. A fraction is in its lowest terms when its numerator 
and denominator do not have any common factor, except unity. 

Thus : y, y, x y are in their lowest terms. 

6 b x — y 

109. The first fundamental principle of fractions. 

If both terms of - 2 ^ be divided by 4, f is obtained. - 2 ^ = 3, 
and also f = 3. The value of - 2 ^, therefore, is not changed. 


Rule. — If the numerator and the denominator of a fraction 
are both divided by the same number, the value of the fraction 
is not changed. 


Example 1. 


Consider the fraction 


(x ~ y)(x + y) 
(x - y){x - y) 


If the numerator and denominator are both divided by x — y, the 
resulting fraction is 

x - y 

Let x = 3 and y — 1. 

Then (*-y)(*+y) = ( 3 ~ 1 )( 3 + 1 ) = 2li, or 2. 

(x-y)(x - y ) ~ . . ° 

(x + y) 

(x - y) 

... ( x ~~ v)( x v\ does equal x + when x = 3, and y — 1. 

(x - y)(x - y) x - y 

They will be found equal for all values of x and y except when x = y. 

139 


(3 — 1)(3 — 1) 2-2’ 


Also 


3 +1 = % or 2. 
3-1 2’ 











140 


ALGEBRA 


110. Rule. — To reduce a fraction to lowest terms: 

1. Factor completely the numerator and denominator. (§95) 

2. Divide numerator and denominator by all their common 
factors, and write the quotients, including any l’s obtained as 
quotients. 

3. For the numerator of the result, take the product of the 
numerator quotients; for the denominator, the product of the 
denominator quotients. Check by substitution. 

Example 2. Reduce ^ m m—168 j owes ^ terms. 

ra 2 - m - 42 


Solution. 1. 


2 . 


3 ra 2 + 3 m — 168 _ 3 (ra 2 + m — 56) 
m 2 — m — 42 (ra — 7)(ra + 6) 

1 

_ 3(ra + 8)(m-— '-f) or 
(ra—=-T) (ra + 6) ’ 

1 


3(ra + 8) 
.ra + 6 


Check. Let ra = 5. 

m L 

Also, 3(w + 8) - 3 < 5 + - 8 -> 


3 ra 2 + 3 ra - 168 _ 75 + 15 - 168 Qr 39. 


ra + 6 


5 + 6 


ra - 

11 


42 25 - 5 - 42 7 11 

.*. the result is correct. 


Note. — The process of crossing out common factors from numera¬ 
tor and denominator has long been called cancellation. The word is 
unnecessary. 


A serious error in reducing fractions is illustrated by the 
solution below, written by a pupil. 


1 

2 a + h _ + b 2 + 5 

ac fic ’ c 

To prove that the result is wrong, let a = 2, h = 4, 

2«+6 = i±i = 8 2i6 = 2+4 = 

ac 6 6 c 3 


and c = 3. 

l’ or2 - 


The error occurred when a of the numerator was divided by a. 

The numerator can be divided only by factors of the numera¬ 
tor, and a is not a factor of the numerator. It is a factor only 
of one term of the numerator. Never cancel terms or factors 
of only a term of a numerator or a denominator. 


















FRACTIONS 


141 


EXERCISE 80 

Reduce to lowest terms : 


16 

. 7 m 2 

4. - 

^ 12 a 2 b 2 c?(x — y) 

30 

16 mn 

18 ab 2 c(x — y) 

24 

15 r 2 s 

8 26 m(x + y)(x — y) 

44 

24 rs 2 

36 m 2 {x + y) 

32 

6 12 A/ 2 

9 0 - s)0 + 2 s) 

72 

34 x 2 )/ 4 

( r — s) 2 (r + s) 

10. 

(« — 2 ?/) 2 

21 * x 2 + 3 x — 18 

x 2 — 4 y 2 

X 2 - 8 X + 15 

11. 

3 a: 2 + 12 

n „ * a 2 - 7 a - 18 

a: 4 — 16 

a 2 -J- 3 a 2 

12. 

x 3 — a;?/ 2 

* 5 c 2 + 15 c - 20 

4 x 2 4 xy 

23 ' c 2 + 6 c — 7 


a 4 - 9 

* 2 x 2 — 2 x — 24 

x o. 

(a 2 - 3) 2 

5 a: 2 — 25 x + 20 

14. 

5a + 10 

* ax 2 — 3 axy + 2 ay' 2 


a 2 — 4 

6x 2 — by 2 

15. 

4 m 12 

og * m 2 — 7 mn + 12 n 2 


2 TO 2 - 18 

m 2 — 9 mn + 20 ra 2 

1 A 

3 c 3 — 12 raft 2 

* 5 x 2 — 45 1 / 2 

lb. 

5 c 2 — 10 cd 

4 a: 2 + 4 xy — 24 ?/ 2 

1 *7 

a + b + c 

* 6 a 2 - a - 1 

11. 

(a + b) 2 - c 2 

‘ 6 a 2 - 5 a + 1 

18. 

0 + 2/) 2 — 1 

* 8 c 2 — 10 cd — 3 d? 


3 x + 3 y + 3 

12 c? — 5 cd — 2 cP 

1 Q 

3 a 3 — 3 ab 2 

30 * x* — y* 

xy. 

(3 a 2 + 3 aft) (a — 6) 

x 2 — 2 xy + y 2 

Ofk 

(2 a + 2 ft) (a + 2 ft) 2 

31* * ~ yi 

zu. 

(5 a 2 — 5 b 2 )(a + 2 6) 

a: 4 + 2 x 2 y 2 + y* 





























142 


ALGEBRA 


MULTIPLICATION OF FRACTIONS 


111 . The product of two fractions is the product of their 
numerators divided by the product of their denominators. 


Example 1. 


5 a 
2x 


3 x 2 _ 15 ax 2 
10 a 2 20 a 2 x 


15 ax 1 

Reducing to lowest terms: 20 a 2 x 


3-lx 

- i **■<”* ' or 3 x 
4 a 

4 a 1 


However, before multiplying, it is preferable to remove any factors 
common to a numerator and a denominator by dividing both the nu¬ 
merator and denominator by such common factors. 


Thus, 


5 a ' 3 x 2 _ m 3 ;r Qr 3jc 
2 x 10 a 2 2/ 1^/ 4a 
2 a 


Rule. — To find the product of two or more fractions : 

1. Find all of the prime factors of the numerators and de¬ 
nominators. 

2. Remove factors common to a numerator and a denomina¬ 
tor by dividing the numerator and denominator by such com¬ 
mon factors. 

3. Multiply the remaining factors of the numerators for the 
numerator of the product, and of the denominators for the de¬ 
nominator of the product. Check by numerical substitution. 

5 a - 30 2 a 3 - 4 a 2 


Example 2. Find 
Solution. 1. 


4 a(a - 2) 2 a 2 - 36 


1 1 a 1 

5 a — 30 2 a 3 — 4 a 2 _ 5(a—HI) ^2) 

4 a(a - 2) 2 ‘ a 2 - 36 2) (a - 2) * (a-—6)(a + 6) 

2-11 1 
2. = 5a 

2(a - 2) (a + 6)’ 

Note, a cannot be 2, or 6 in this example, for when a = 2 (a — 2) 2 
= 0, and the first fraction does not have any meaning; when a = 6, 
the second fraction does not have meaning. 















FRACTIONS 


143 


EXERCISE 81 

Find the following indicated products : 

1 . 

2 . 2 

8 

3. ^ 


16. 


17. 


18. 


19. 


20 . 


21 . 


22 . 


5 14 

6. 

3 

14 11 

11. 

2 xy 2 3 xy 

6 5 


10 

’ 33 ' 7 


~3x? ' 

3 20 

7. 

4 

15 7 

12. 

4 a 3 6 15 x 2 y 

8 ’ 15 


5 * 

14 * 6 


5 x 2 4 a 2 b 

5 6 

Q 

6 

14 iO 


5 a 2 6 be 2 

2 ' 33 

8. 

21 

5 4 

13. 

4 6c 10 a 2 6c 

3 a 2 b 

9. 

2 a 

6 6 5 c 


27 x 3 y 2 30 xz 2 

4:b 5 a 


3 b 

7 c 4 a 


20 x 4 z 18 y 3 

6 x 10 y 

10. 

2 ab 3 be 5 ac 

15. 

10 mV 3 rfir 3 

~5y 2 ' 3T 2 

3d 

! 4 a 2 6 6 2 

6 nr 3 5 mV 

x + y 3 x 2 



23. 2 0,2 

— a 

\ 6 a -f - 3 


a 2 — b 2 5 a 3 


4 a 2 (a -j- b) 2 
6 m + 12 5m — 


15 


3m - 9 

4 m + 8 

am + ax 

cm — cx 

dm — dx 

bm + bx 

4 m 2 - 1 

m + 4 

m 2 — 16 

2 m 1 

7 m + 14 

m 2 — 4 

3m - 6 

(m + 2) 2 

xr + xs 

mr — ms 

yr - ys 

nr + ns 

30 ® 4 

- 6 4 4 a 5 

3 a 

- 3 6 (a 2 


24. 


25. 


26. 


27. 


28. 


29. 


4 a 2 — 1 

4 a 

1 — 

00 

1 

3 z — 6 

4 + 4 x 

5 m + 1C 

> 4m - 2 

8m - 4 

3 m + 6 

a 2 - 6 2 

ac + 6c 

(a + &) 2 

ac — 6c 

2« + 8 

51-10 

t 2 - 4 

51+20 

a; 2 - 16 

3 a: 2 — 6- x 

0 - 2) 2 

2 x 2 + 8 x 

CO 

1 

Cn 

to 

1 

CO 

to 

2 2/ + 8 

[IO 

t- 

1 

% 

CO 


■ b 2 6 a 2 + 6 6 2 


(a + 6) (4 a -f~ 2 6) 


Note. — Examples involving optional factoring appear on page 144. 







































144 


ALGEBRA 


EXERCISE 82* 

(This exercise involves optional factoring. See § 97, § 98.) 

2 2 + 8 52 - 10 3t/-15 ^ + 92/ + 8 

* 2 2 — 4 ' 2 2 + 6 2 + 8 ‘ 2 y+ 8 y 2 + 3 y - 40 

3 x 2 - y 2 m 4 x - 8y 
x 2 — 4 y 2 3 x 2 + 3 xy — 6 y 2 

^ a 2 x — a 3 _ x 2 — ax — 6 a 2 
x 3 — 3 ax 2 x 2 + ax — 2 a 2 

a 2 - 6 a - 27 a 2 + 10 a + 24 
' a 2 + 9 a + 18 ’ a 2 - 7 a - 18 

2 r 2 + 5 rs — 12 s 2 r 2 - 2 rs + s 2 

2 r 2 — 5 rs + 3 s 2 r 2 + 5 rs + 4 s 2 

3a— 66 3 a 2 2 ab — b 2 a 2 — ab — 6 fe 2 

9 a 2 — b 2 a 2 — 4: b 2 6 a + 6 b 

a 4 — ¥ 4 a 2 — b 2 6 a 2 + 6 b 2 

3a — 3 b a 4 + 2 a 2 b 2 + 6 4 2 a 2 + ab — b 2 

9 4 x 2 + 12 x + 9 3 g 2 +11 x - 20 \ 3 z 2 + 22 a; + 7 

3 £ 2 + 17 £ — 28 2 a: 2 - x - 6 ’ 2 z 2 + 13 x + 15 

3 m 2 — m ' 2 m 2 + 15 m + 25 6 m 2 — 15 m 

4 m 2 — 25 6 m 3 m 2 + 14 m — 5 

u.. 4 ~* 2 . 20 -*- . 6 ~ 2 * 

12 — 3 a: 4 — 4x + x 2 10 + 7 a; + z 2 

^2 a a; 2 — mx + m 2 _ 2 x 2 — 2 mx x 2 — 2 mx — 3 m 2 

2 x 2 — 2 m 2 5 x 2 — 15 m.r 3 a: 2 — 3 mx + 3 m 2 
b. What is the value of the indicated product in part a when 
x = 3 and m = 2 ? 

4 a 2 — 25 y 2 t 15 x 2 - 55 xy - 20 y 2 2 x - 4 y 

10 x — 20 y 2 x 2 — 3 xy — 20 y 2 6 x 2 — 13 xy — 5 y 2 


13. 


































FRACTIONS 


145 


DIVISION OF FRACTIONS 


112. In arithmetic , to divide one fraction by another, we 
multiply the dividend by the inverted divisor. 

3 

10 ' 5 M $ 2 

2 1 

1 

11 v 5 
3*5 3 

1 

In algebra, we do the same. 


Thus, a. 


b. 


3i*2* = iUll=^X^or! 


Rule. — To divide one fraction by another: 

1. Factor the numerators and denominators of the fractions. 

2. Invert the divisor fraction. 

3. Multiply the dividend fraction by the inverted divisor. 
Note. — The divisor follows the sign of division. 


Example. Divide + 10 <T + 21 ™ by “ V ~ 9 m * 


4 a 2 + 3 a 


ar — or 


Solution. 


2 . 


3. 


4. 


j a?m + 10 am +21 m a?m? — 9 m 3 
a 3 — 4 a 2 + 3 a a? — a? 

_ m(a? + 10 a + 21) m 3 (a 2 — 9) 

a(a 2 — 4 a + 3) a 2 (a — 1) 

1 1 a 1 

_ jm(a + 7)(a—K3) . — ^T) 

^r(a -- 3) (a—"+) ;m+a — 3) («—h"3) 

1 1 m 2 1 

= a(a + 7) 
m 2 (a — 3) 2 


Check. Let a = 2 ; m = 1. 

a 2 m + 10 am +21 m . a 2 ra 3 — 9 m 3 _ 4 • 1 +10 • 2 • 1+ 21 4 — 9 

o 3 — 4 <z 2 + 3 a * a 3 - a 2 8-16+6 *8-4 

1 1 

a(a + 7) _ 2(2 + 7) = 2-9 = 18 = 18 
m\a- 3) 2 1(2 -3)* 1(-l) 2 1 


Also, 


















146 


ALGEBRA 


1 . — -f- 




EXERCISE 83 


3 . 2 

3 

•1 

00 

3 X 

36 m 2 9 m 3 

14 ’ 7 


5 y 2 

10 y 

c 

col 

Oi 

Sfj 

9a. 15 a 

4. 

12 c 2 d . 

4 d 

6. 24 ary -r- 

-o 

00 

rO 


28 cd 2 ' 

’ 7 c 

* 4 

7 2c - 

4 c 2 


10. 

12 - 4 x 9 - x 2 

' 4 c 2 - 1 

6c+3 

10 + 5 x ’ 4 - x 2 

o 5a + 15 

. 5a — 

15 

11. 

x 2 — xy . x 2 — j/ 2 

' a 2 — 9 

(a - 

3) 2 

3 x — 6y 3(x + y) 2 

9 + 

c 4 - d* 


12. 

a 3 + ab 2 . a 2 + 6 2 

* C+d 

(c - d) 

2 

a 3 — 2 a?b (a - 2 by 



EXERCISE 84* 



(These examples involve optional factoring.) 

1 3x-\-Q^_x 2 — 2x — 8 
x 2 - 4 * x 2 + x- 12 

2 8-4 y 10-7 y + y 2 

' 12 + 62 / ' 10 + 3 y - y 2 

g x 2 — 2 xy a : 2 — 3 xy -(- 2 y 2 

xy + 3 y 2 x 2 + 4 xy + 3 y 2 

4 a 2 + b 2 6 a 3 + 6 ab 2 ^ r 2 — rs r 2 — 2 rs s 2 
a — b a 2 — 2 ab + b 2 r — 2 s r 2 — rs — 2 s 2 

g 4 r 2 + 4 rs + s 2 4 r 3 — rs 2 
8 r — 4 5 8 rs — 4 s 2 

^ x 2 — x — 42 x 2 2 x — 8 

x 2 + 5 x - 84 ' a; 2 + 10 * - 24 
g m 3 — 5 m 2 — 6 m 2 m 2 — 14 m + 12 
3 m 2 + 9 m + 6 ra 2 + ra — 2 
g c 2 — 5 cc/ + 6 c? 2 . c 2 + 2 c<i — 15 <i 2 

c 2 - 9 cd + 14 d 2 * <? -2 cd -35 d 2 

10 12 a 2 - 17 ab - 5 b 2 . 8 a 2 - 2 ab - b 2 

9 a 2 - 30 a6 + 25 6 2 * 9 a 2 - 18 a& + 5 b 2 

a 2 - 16 a 2 — 16 a + 55 3 a - 12 

2a 2 — 9a—5 a 2 — 7a — 44 ‘ 4a 2 - 1 


11. 






































FRACTIONS 


147 


ADDITION AND SUBTRACTION OF FRACTIONS 

113. In arithmetic, fractions having a common denominator 
are added or subtracted without difficulty: 

Thus, a. ts- + tz ~ T¥- h. A — tb = ts- 

Fractions which do not have the same denominator must be 
changed to equal fractions which have a common denominator. 

Thus, fH - T _ l _ i = A'"hA - l - A' or x¥* 

12, the lowest common denominator , is the smallest number which 
contains 3, 2, and 4 exactly as divisors. While any common denomi¬ 
nator can be used, it is natural to use the lowest common denominator. 

In algebra, also, before fractions can be added or subtracted, 
they must be changed to equivalent fractions having a common 
denominator. 

You must, therefore, learn: 

a. how to form the lowest common denominator; 

b. how to change fractions to equivalent fractions having 
this common denominator. 

114. Lowest common multiple. 

a. A multiple of a number is any number which contains the 
given number as an exact divisor. 

Thus, 6, 9, 12, and 30 are multiples of 3. 

b. A common multiple of two or more numbers is a multiple 

of each of them; it can be divided by each of them. 

Thus, 24 is a common multiple of 2, 3, 8, and 12. 

Also, 48, 72, 96, and 120, are common multiples of 2, 3, 8, and 12. 

c. The lowest common multiple of two or more numbers is 
the smallest of their common multiples. 

Thus, 24 is the lowest common multiple of 2, 3, 8, and 12. 

' d. The lowest common multiple of two or more expressions 
is that multiple of them which has the least number of prime 
factors. 


148 


ALGEBRA 


Example 1. What is the lowest common multiple of 5 a 3 b 
and 2 ab 5 c ? 

Solution. 1. The smallest number which will contain both 5 and 
2 is 10, — namely, 5-2. 

2. The lowest power of a which will contain a 3 and a (separately) is 
a 3 ; that is, a 3 -s- a 3 = 1; a 3 -f- a = a 2 . 

3. The lowest power of b which will contain b and b h is 6 6 . 

4. The lowest power of c which will contain c is c. 

5. the L. C. M. = 10 a 3 b h c. 

Check. Does 10 a 3 b 5 c contain each of the expressions? 

Yes, for 10 a ^~ =2 6 4 c; and ^ a ^~ = 5 a 2 . 

5 a 3 b 2 ab 5 c 

Rule. — To find the L. C. M. of two or more expressions: 

1. Find the prime factors of each of the expressions. 

2. Select all of the different prime factors and give to each 
the highest exponent with which it appears in the given expres¬ 
sions. 

3. Form the product of all of the factors selected in Step 2. 

Example 2. Find the L. C. M. of 2 am 2 (m -j- a) and 
9 a 2 (m + a) 3 . 

Solution. 1. 2 am 2 (jn + a) = 2 • a • m 2 • (m + a). 

9 a 2 (m + a) 3 = 3 2 • a 2 • (m + a) 3 . 

2. 2 appears with 1 as its highest exponent. 

3 appears with 2 as its highest exponent. 
a appears with 2 as its highest exponent. 

m appears with 2 as its highest exponent. 

(m + a) appears with 3 as its highest exponent. 

3. /. L. C. M. = 2 • 3 2 • a 2 • m 2 • (m + a) 3 

= 18 a 2 m 2 (m + a) 3 . 

Note. — Do Step 2 mentally; write only Steps 1 and 3. 

Check. Does 18 a 2 w 2 (m + a) 3 contain each of the given expressions? 

Yes, for = 9 a(m + o)» 

2 am?(m + a) 

18 a 2 m 2 (m + a) 3 _ 2 2 

9 a 2 (w + a) 3 


and 






FRACTIONS 


149 


EXERCISE 85 


Find the L. C. M. in each example, and check: 


8. 4 r 2 and 14 r 3 

9. 16 r 4 and 12 rs 3 

10. 30 m 3 and 18 mn 4 

11 . 5 a 2 , 10 b 2 , and 15 c 2 

12. 4 c 3 , 6 c 2 d, and 10 ccP 

13. 20 x, 15 xy, and 12 x 2 y 

14. 12 y 3 , 8 x 2 y, and 15 x 3 


1. 6 and 8 

2. 12 and 15 

3. 18 and 12 

4. 6 a and 9 b 

6. 4 x 2 and 10 y 2 

6. 4 ab and 6 a 2 b 

7. 8 mn and 12 n 2 


15. 27 c 4 , 6 c 3 d, and 18 c 2 d 2 

16. (x + 2)(x + 5) and (x + 2)(x — 3) 

17. (m — 2 n ) 2 and (to + w) 

18. (1 + a) 3 , (1 — a)(l + a), and (1 + a) 2 

19. 3 (a + b) 2 , 4 (a — b) (a + b), and 2 (a — b) 2 

20. 2 x — 10 and 3 x — 12 

21. 3 a — 7 and 6 a + 15 

22. 5 a + 5 b and a 2 — b 2 

23. 4 f — 1 and 2 p + 1 

24. x 2 — y 1 and (x — y) 2 

25. mx — my, nx + ny, and x 2 — y 2 

26. z 2 — 9 and (x + 3)(x — 11) 

27. 4 c 2 — 25, 6 c — 15, and 8 c + 20 

28. * a 2 + 2 ab + b 2 and 2 a 2 + ab — b 2 

29. * 4 a 2 — 9 b 2 and 4 a 2 + 12 ab + 9 b 2 

30. * 9 mx 2 — m, 3 mnx — mn, and 9 nx 2 — 6 nx + n 

31. * x 2 — 10 x + 21, x 2 — 3 x — 28, and x 2 + x — 12 
32 * 6 r 2 + 11 rs — 10 s 2 and 12 r 2 + rs — 6 s 2 

33. * 10 x 2 — x — 21, 15 x 2 + x — 28, and 6 x 2 — 17 x + 12 

34. * 6 a 2 — a — 1, 3 a 2 + 7 a + 2, and 2 a 2 + 3 a — 2 


150 


ALGEBRA 


115. The second fundamental principle of fractions. 

If both terms of the fraction f are multiplied by 4, the result 
is Since both fractions have the value 3, they are equal. 

Rule. — If the numerator and denominator of a fraction are 
both multiplied by the same number, the value of the fraction 
is not changed. 

This principle is used in changing fractions to a common 
denominator. 

116. The lowest common denominator (L. C. D.) of two or 
more fractions is the lowest common multiple of their de¬ 
nominators. 

Example 1. Change to their lowest common denominator 
the fractions f and f. 

Solution. 1. The L. C. D. of these fractions is 12, for 12 is the 
L. C. M. of 4 and 6. (§114.) 

2. To change the denominator 4 into 12, it must be multiplied by 3; 
that is, by 12 -f- 4. Then, in order not to change the value of the 
fraction, multiply also the numerator by 3. 

. 3 = 3j_3 = _9. 

"4.3-4 12* 

3. For the fraction £: 12 .-f- 6 = 2. Multiply numerator and de¬ 
nominator by 2. 5 _ 2 • 5 _ 10 

" 6 ~~ 2^6 ~~ 12 * 

The fractions have been changed to ^ and ff. These fractions 
have the common denominator 12. They have the same value as the 
fractions £ and f, because they were derived by use of the rule in § 115. 

In algebra, use the following 

Rule. — To change fractions to their lowest common de¬ 
nominator : 

1. Find the prime factors of the denominators. 

2. Find the L. C. M. of the given denominators; this is the 
L. C. D. 

3. For each fraction, divide the L. C. D. by the given denomina¬ 
tor ; multiply both numerator and denominator by the quotient. 


FRACTIONS 


151 


Example 2. 


Change to their lowest common denominator: 
4 a , 3 a 


a 2 — 4 


and 


5 a -fi 6 


Solution. 

i. f a - 

4 q 

: -—— • 



a? — 4 

(q — 2) (q + 2) 


2. 

3 a 

3 q 


q 2 — 5 a + 6 

(a — 2)(q — 3) 


3. 

The L. C. D. is 

(q + 2)(q — 2) (a — 

3) 

4. 

L. C. D. -r (q 

- 2) (a + 2) = a - 

3. 


. _4 a __ _ 4 q(q — 3) _ 

(a — 2) (a + 2) (a — 2) (a + 2) (a — 3) 

5. L. C. D. -5- (a — 2)(a - 3) = (a + 2). 

. _3a___ 3 a(a + 2) _. 

(a - 2) (a - 3) (a - 2) (a + 2) (a - 3)* 


Check. — The final fractions in Steps 4 and 5 may be changed into 
the original fractions by dividing out common factors. 

Note 1 . — In Step 4, to get L. C. D. + (a — 2)(a + 2), write, if 
necessary: 

- 3) = a _ 3. 


(a- 2) (<H-2) 

Similarly, in Step 5: 

(a + 2) -2) (a-3) 


(a—-2) (a—--3) 

Try to do this mentally, however. 

Note 2. — In Step 4, the fractions 
4 a(q — 3) 


q + 2. 


4 q 


(q - 2) (q + 2) 


and 


(a - 2) (a + 2) (a - 3) 


are “equal.” 


Really they are identically equal. (See § 75.) Such fractions are 
called equivalent fractions, to distinguish them from fractions (which 
will be studied soon) which are equal for some few values of their letters. 

when x = 2 but not otherwise. 


Thus, ? = 

* 2 3 


Remember: equivalent fractions always have the same value but 
differ in form. 

















152 


ALGEBRA 


EXERCISE 86 

Change to equivalent fractions having their L. C. D.: 


1. 

5. 

4 

6. 

4 a; 

2 

3 2/ 2 

11. 

2. 

4. 6 



8’ 

3 

T 


2 


a ’ 

b ’ c 


2. 

3 

3 


3. 

4 


12. 

2 a 

3 6 

4 c 

5 ’ 

4 

7. 

> 

o: 

y 


m 

’ m 2 ’ 

m 3 


a 

a 

8. 

7_. 

8 


13. 

a m 

6 c 


3. 

6 ’ 

4 

m ’ 

71 


X 

a 2 ’ a: 3 


A 

m _ 

n 

q 

9_. 

4 

* 

14 

5 

3 

_£ 

4. 

8 5 

5 


O j 

X 1 

X 



2x 

4 y 

3 a 


2 a 

3 6 

10 

6 . 

7 


15 

2 a 

. 36. 

4 c 

o. 

3 

’ 4 

Iv. 

ah' 

b 



mn 

nr 

mr 


16. 

17. 

18. 

19. 

20 . 


x 


y 


x + y x - y 
a b 


2a — b’ 2 a -\- b 
2 m 3 n 


3m +w’ 2m — n 

_3_. 4 

2 r — 6 ’ 3 r — 9 
a b 


5x + 10’ 7 x - 21 

* 3 x 


26. 

27. 

28. 

29. 

30. 


21 . 

22 . 

23. 

24. 

25. 

5 x 


x 2 — 9 ’ x + 3 
3 2 


4m 2 — 1 ’ 6m + 3 
3 r 2 r + s 
r 2 — 25 s 2 ’ r — 5 s 
x 2 — y 2 . x + 2 y 
(2 x - 3 y) 2 ’ 10 x - 15 y 
m n 

(2 m — 3 n) 2 ’ 4m 2 -9n 2 


z 2 — 4 x - 21 ’ z 2 — 2 x — 35 
^ 3 a 2 a 


4 a 


6 a 2 — a — 

1 ’ 3 a 2 + 7 a + 2 ’ 

2 a 2 -f 3 a — 2 

m + 1 


m — 4 

m + 3 

m 2 — m — 

6’ m 2 

— 4 m + 3 

m 2 + m — 2 

r 


a 

ar 

r 2 — 6 ar + 9 a 2 ’ 

r 2 + 4 ar — 21 a 2 ’ r 2 + 7 ar 

2 

1 

3 



Sx — 6 y' 2 x —3 y* 8 x — 4 y 




































FRACTIONS 


153 


117. Adding and subtracting fractions is now possible. 

Rule. — To add or subtract fractions: 

1. Reduce them, if necessary, to equivalent fractions having 
their lowest common denominator. 

2. For the numerator of the result, combine the numerators 
of the resulting fractions, in parentheses, preceding each by 
the sign of its fraction. 

3. For the denominator of the result, write the L. C. D. 

4. Simplify the numerator by removing parentheses and 
combining like terms. 

5. Reduce the result to lowest terms. 


Example. Find 
Solution. 1. 

2 . 


5z - 4y 7 x - 2 y ^ 
6 14 

5 x — 4y _ 7 x — 2 y 
6 14 


= 7(5 a; - 4 y) _ 3(7 x - 2_ y ) (Rule> Step ^ 
42 42 


3. 


7(5 x-4 y) -3(7 a:-2 y) 
42 


4. 


35 x - 

-28 y — 21 * + 6 y 



42 

5. 

= 


14 x — 22 y 

42 

6. 

= 

2(7 x- 11 y) 

42 

7. 

= 


7 x — 11 y_ 

21 

Check. 

Let x = 1; 

y=l. 

5 x — 

4y_5-4 or 

1. ] 

1 and I_ 5 = 7 _ 

6 

6 

6’ 

6 14 42 

7 x — 

2j/_ 7 - 2 or 

5 . 

Also, 7x ~ ny 

14 

14 

14 ’ 1 

1 ’ 21 


(Rule, Steps 2,3) 
(Rule, Step 4) 
(Rule, Step 4) 

% 

(Rule, Step 5) 


15 


= or ^f 

- 11 ^ 4 

21 ’ 21 


Note. — After a few examples, omit Step 2; pass at once from 
Step 1 to Step 3. Always write Step 3. Notice the words “in paren¬ 
theses” in the second step of the rule. 




















154 


ALGEBRA 


EXERCISE 87 

Simplify, patterning after the solution on page 153: 


i.M 

10. 

2 

3 5 

19/^ + 1 + g-1 

4 6 


3 

4*6 

2 4 

2.1-1- 

11. 

2 

L + I 

20 y^A~y-±A 

5 15 


5 

15 3 

3 6 

3 J-? 

12. 

a 

a .a 

nl 4a—3 . 2 a + 1 

8 3 


8 

2 ~4 

8 6 

. m . m 

4.--- 

13. 

V - + 

y y 

oo « + 3 , 3 a — 2 

5 10 


6 

4 8 

3 5 

5. — 

14. 

2x 

z , 5 a; 

03 x -3y . 2z + 3y 

12 4 


3 

2^6 

4 14 

. 5 a . a 

o.-r — 

15. 

2t 

9t . t 

5c + 4 c + 3 

9 3 


7 " 

14 + 2 

12 9 

5m ,3m 

16. 

3 w 

,3 w 3 w 

25 2 c - 1 _ c + 2 

""6 5~ 


~ST 

+ T" IT 

5 6 

8 3s 2x 

17. 

a - + 

b , c 

_. 3m + 7 2m — 1 

Zo. 

2 3 


3 

4 + 2 

5 4 

9 13 y 5 y 

18. 

X 

y .2 

4z - 3y 5z -4y 

15 6 


3 

5 6 

6 8 

os 2 z + 5 | a: - 7 5 a; + 2 


2 g 3 a + 5 _ 5 a + 3 _j_ a — 6 


30. 


4 6 

4 £ + 1 6 z — 1 


4x + 3 


6c — 4d 5 c — d . 4 c -\~3 d 
3 2 ‘ ' 5 

32 ^ m + 2 n _j_ 5 m — 2 n _4m -\- n 


33 . ? - r - ~ * - g.. r + 2 * + lr + 3 * _ jr -_4g 


5 


10 


15 


20 































FRACTIONS 

155 

34. Find 

2a + b&2b—c 2c + a 
ab ' be ac 


Solution. 

1 2a + 6|_2b— c 2c + a 

ab be ac 


2. 

_ c(2 a + b) j a(2 b — c) 6(2 c + a) 
abc abc abc 

(Rule 3, 
page 153) 

3. 

_ (2 ac + be) + (2 ab — ac) — (2 be + ab) 

abc 


4. 

_ 2 ac be + 2 ab — ac — 2 be — ab 


abc 


5. 

_ ac — be + ab 
abc 


Check. 

If a = 1, b = 2, and c = 1: 



2 a -{- b | 2 b 


2 c + a 


ab 


be 


2 +2 + 4 


= 2 +! 

be + ab _ 1 • 1 - 2 • 1 + 1 • 2 


3,° r |, 


2 + 1 
1 


2+2 


35. 

36. 


abc 


1-2-1 


2 ’ 2 

—+— 

5 a 3 a 

43. 

4* 3* 

51. 

J_ 2_ _ 3 

2 x 3 x x 

_7_4_ 

2 m 5 m 

44. 

2 3 

5 a 10 a 

52. 

2 __3 4 
a: 3 a: 2 x 

-+§ 

45. 

4 _ 3 
c d 

53. 

a _^_b c 
x y z 

6 _ 7 
r 5 

46. 

c d 
x y 

54. 

2 3,4 

a 2 ab b 2 

a _l_ b 
x a; 

47. 

5 x 4 y 

55. 

J_2_, _5_ 

2 a; 2 3a:y 6 ?/ 2 

i ^ 

1 

CO 1 W 

48. 

a , b 
3*7* 

56. 

2a; - 1 j 3a; + 2 
3 a; ' 4 a; 


41. ? + ? 

X y 

42. —-? 


3 r 


m . n 

49. — H- 

rx ry 

60 x 2 x 3 


57. 


58. 


5 & + 3 y _ 2 # + 4 y 


x 2 y 


xy‘ 


x + 1 a: 2 — 1 


2 a: 


3 a; 5 





























156 


ALGEBRA 


1. Find 


Solution. 1. 


EXERCISE 88 
3 4 


2 p — 6 3 p — 12 


2 p — 6 3 p — 12 

3 4 


( Factoring the\ 
denominators / 

g _ 9(p — 4) _ 8(p — 3) /Step 1, ruleA 

6(p - 3)(p - 4) 6(p -3)(p - 4) \page 153 / 

(Complete the solution, following Steps 2-5 of the rule.) 


Note. — In Step 3, to change 


2 (p - 3) 

a. 6(p - 3)(p — 4) + 2(p - 3) = 3(p - 4). 


to the L. C. D.: 


b. Multiplying both terms of 


we get 


9 (p ~ 4) 

6(p -3)(p-4) 


2 (p - 3) 


by 3(p - 4), 


2 . - -—7 + 1 2 


m — 1 to + 1 

1 _ 1 

TO — 1 TO -f 1 

1 _ 1 

a — 6 a + b 


6 .- 2 -+_ 3 _ 
* + y X - y 

6. :r ———t + ■ 3 


3z - 1 2 # + 1 


11 . 

12 . 

13 . 

14 . 

16 . 


3 a — 7 9 a + 5 

a _ b 
a — b a + b 
6 2 


5y - 3 4y — 1 

2 3 


4 x — 3 a 6 x + a 
4 3 


3 x — 3 y 2 x + 2 y 


7. 

TO 1 

16. 

►o 

H 

CO 

TO — 1 

TO + 1 

5 a + 10 3 a — 3 

8. 

r , r 

17. 

1 5 

r + 3 

r - 3 

Ox — 2y 30 x — 9 y 

9. 

c 

c 

18. 

x , y 

c - 2 

c -}- 3 

2x + 2y 3x - 3 y 

10. 

* I y 

19. 

3b 5 b 

X + y x - y 

3 ci 4 5o + 6 
















































FRACTIONS 


157 


20. Find 

Solution. 1. 


a + b a — b 

4 a 2 - 9 b 2 (2 a + 3 6) 2 ’ 
a + b _ a — b 
4 a 2 -9 b 2 (2 a + 3 6) 2 


__ a -j- b __ a — b 

“(2 a - 3 6) (2 a + 3 6) (2 a + 3 6) (2 a + 3 b) 

_ (a + b)(2 a + 3 b) - (a - 6)(2 a - 3 6) 

(2a - 3 6)(2a + 3 6)(2a +3 6) 

_ (2 a 2 + 5 afr + 3 fr 2 ) - (2 a 2 - 5 ab + 3 fr 2 ) 

(2 a - 3 6) (2 a + 3 6) (2 a + 3 b) 


(Complete this solution and check it.) 


Note 1. — Steps 1, 2, and 3 of the rule on page 153 have been taken 
at once in Step 3 of this solution. The first fraction of Step 2 is equiva- 

lent to fa +fe)(2o + ,3b) . the second to fa -b)(2 a - 3 b) 

L. C. D. L. C. D. 

Note 2. — In Step 4, (2 a 2 + 5 ab + 3 6 2 ) is (a + 6) (2 a + 3 b), 
and (2 a 2 — 5 ab + 3 6 2 ) is (a — 6) (2 a — 3 6). Always inclose such 
products in parentheses as in this Step 4. 


21 . 

22 . 

23. 

24. 


25. 


a — 2 

a + 2 

a — 3 

a + 3 

x — 2 

z + 2 

(s + 5) 2 

1 

3 r + ^ 

9 r 2 + s 2 

CO 

-$ 

1 

CO 

9 r 2 — s 2 

2 m — 3 

. m — 9 

4m - 2 

3 m + 6 

5 £ — 4 y 3 x + 4 y 

5x — y 

3 x + y 


r + 4 _ r + 6 


1 

CO 

1 

n. 

r 2 - 

r — 6 

3 


2 

4 a 2 - 9 

6 a 2 

— 9 a 

X 


X 

x 2 — 6 x + 8 

x 2 - 16 

a + b 

a 

- b 

2 a + 6 b 

3 a 

+ 36 

a + b 


a — b 


x 2 — 2 xy + y 2 x 2 — y 2 


Q1* a + 5 I a + 3 

a 2 -a-6 a 2 +7a+10 

32 * d + 3b _ a - 3b 
‘ a 2 - 7 + 12 6 2 a 2 - ab - 12 6 2 




































158 


ALGEBRA 


MIXED EXPRESSIONS 


118. An integral expression is an expression which does not 
have literal numbers in the denominators of fractions; as, 
a 2 + b 2 , or fa + b. 

An integral expression may be considered a fraction with 
denominator 1; thus, a + b is the same as — • 

119. A mixed expression consists of one or more integral 
terms plus or minus one or more fractional terms; as fl 2 +-, 


or 1 + 


cl -j - b 


120. Changing mixed expressions to fractions is merely a 

matter of adding or subtracting fractions. 

Thus, 12f = 12 + - = + - 3 = — • 

7 8 8 8 

Also, x + ~ 3 = 3 \~ - 3 - 

Z 1 z z z 


EXERCISE 89 

Change the following mixed expressions to fractions : 


1. 10 


2.5-1 


3.| + 2 

4j-3 

5 


5. cl -b 


•'"3 


’■ 2 -l+J 

*l +2 -5 

9. 1 + - + 5 a 
2 


10 . 2 


5 — 2 x 


11 . a - 1 + — 
2 a 


12 . c 


4 + — 
3c 


13. 1 _2x_-3j/ 
2 x + 3y 

b 2 


14. a + b + 


15. 3 a 

16. 2 + 

17. 6 + 

18. 1 - 


a — b 
9 a 2 


3 cl 4- b 


2s z 


r 3 -s 3 
5 x + 2 

x 2 -1 
a — 16 
















FRACTIONS 


159 


19. 2 + 

1 

23. 

S ' 2 

2 a ; 2 

+ 1 

27. 

1 

2 6 3 

a 


3 

i 

+ 

20 . 3 - 

2 

24. 

n — 

a 2 - 

1 

28. 

1 

2 a& 

b 

it 

a 


a 2 + ab + b 2 

21. x — 

2 

25. 

a 


1 

29. 

3 

a — 5 x 


y 

a — 

6 


a — 2 x 

22 . 2 + 

r 

26. 

a; — 

z 2 


30. 

5 

a 2 — 19 x 2 

O J o 


121. Multiplication and division of mixed expressions. 

In such examples, the mixed expressions are inclosed in paren¬ 
theses. Change the mixed expressions to fractions, doing what 
the signs direct; then multiply or divide as the signs indicate. 


Example. 

Solution. 

2 . 

3. 

4. 

5. 


Fi ” d ( 4 -JTT) + ( 18 + ?^i> 

■- (*-rh 


- /4(s + 1 ) - F\ ^/ 16Q 2 - 1) + 7 


T 


_^ 4(rc + 1) - 1 ) 

= 4x + 4 - 1 ^ 16 x 2 

X + 1 x 2 — 1 

_ 4 % + 3 _ X 2 - 1 
x + 1 

1 

1 

X — 1 


X 2 ~ 1 

16+7 


; ) 


16 x 2 - 9 

1 

(x - l)(^-+-t) 

(4-aH='3)(4a; -3) 

1 


4 x - 3 

Check. Let x = 2. 

- 


( 


( 4 -^) 

*t) 


11 


1 1 

11 . 55_J4T 3 


16 + 


7 _ 


55. 


Also, 


x — 1 
4 * — 3 


= 16 + 4= 18^, or 




T' or 5’ 
1 5 


























160 


ALGEBRA 


EXERCISE 90 


' AlXA) 

‘(—iX-a) 

* (f+’Xf-O 

2 y 




a 2 - 2 6*> 


*e-s(‘+sA-j 


2 6—3 

r 


° aika) 

•AIM 2 -!) 


“•AAM 3 -^) 


a: + 3; 

2 ab + b 2 \ 


i3. (y - i £\+( jL - 8 A 

\x y ) \4 x 2 i/ 2 / 

“ GHMfsA) 


(' + n» - ab + i> ! ) ' 0 + a ! + 6 s ) 

+-D 

”• ( 3 +ferH 3 + A) 

1e /t , 2ai+2a 2 \ / 0 a; + 4 a \ 

18 -0 + **-«* ;a 2 _ 2 x + 2J 






















FRACTIONS 


161 


20 / 


21 / 


8 x 


x 2 + 4 * + 3 

1- 12 y* 

ic 2 +6 xy-\r 5 y‘‘ 


X 2 + / t ) + ( 3 + /- 5 ) 
)( 


!_ 2 xy—4 y 2 y. ^ 2 y ^ 


z 2 -b5 xy— 14 y 2 


x+yJ 


Required by C.E.E.B.; not explicitly referred to in Syllabus. 

122. A complex fraction is a fraction having one or more 
fractions in either or both of its terms; as, 


2 +- 
a 


and 


— 6 a + b 
T I ‘ 

+ 


a — b a + b 


Since a fraction is merely an indicated division, complex 
fractions are merely another way of writing division of mixed 
expressions. a a 

Example. Simplify a ~ b _ a + b # 

b _j_ a 
a — b a + b 

Solution. This means: a. Find the difference in the numerator; 
b. find the sum in the denominator; c. find the quotient. 

The first two of these steps are carried along simultaneously below. 

a _ a a{a + b) — a(a — b) 

l a — b a + b _ (a — 6) (a + b) 

b , a b{a + b) + fl(a — b) 

a — b^~ a + b (a — b)(a + b ) 

1 1 

__ $ ab — pr + ab {a — =rJ b) (a—b~6) 

(a—Hi) (u—Kfr) qS + 6 2 + a 2 — (S 

1 1 

0 _ 2 ab 

a 2 + 6 2 

(Check this solution by letting a = 2 and 6=1.) 

Note. — A complex fraction can be simplified by multiplying its 
numerator and denominator by the L. C. M. of the denominators of the 
fractions in its numerator and denominator. 





























162 


Simplify: 
i. 1 


-i 


1 - 


1 


3. 


6 . 


6 . 


2 

3 

2 + 


‘-fo 

8 +I 

5 -i 

!- 


1 + 

3 

c 


1 


7. 


14 m 


i -i 


ALGEBRA 
EXERCISE 91 


y 


-V 


9. 


10 . 


11 . 


12 . 


13. 


14. 


y 

X + 




2 1 

X Z - 


2 x — 


2x+ 3 


o - y 


+ y 


a ,56 
3 + 1T 


a - 

c 

a , 6 
b a 


17. 


18. 


19. 


20 . 


21 . 


i + “ 

b 


1 _ a , a_ 
b 2 

2 - — 


2 a _ 3_6 

3 6 2 a 

X - 13 + — 
_ 

15 

X 


x — 3 


i _ 1 _ 21 

y y 2 

2 _ 14 

y y 2 

a 


a — b 


1 + 


a — b 

z 

22 . —— 


r 1 

x 2 ~y~ 

1 - * 

O- 

C 

4 

x + y 

7 m 2 n 

, 6 
« H— 

15 c 

2 a i 

no a + 6 

X 

b 

aO. 

i a 


1 - 


24. 


a + 6 
4 s 
r + s 


r_ _ 3_s 

3 s r 


8 . 


16. 



































FRACTIONS 


163 


REVIEW EXERCISE IX 

Perform the indicated operations and check the solution: 
m + 2 m 2 + 2 to 3 - 3 


7 m 14 m 2 21 m 3 
£ 2 + 5 xy -f 6 y 2 # x 2 — 7 xy 
x 2 — 4 xy — 21 y 2 x 2 — 4 ?/ 2 
9 — x 2 6 — 2 x 
5 + x ’ 25 - x 2 

,2 i 


4. («L±r-2mWl-A} 
\ r / \r m/ 

_ a + 6 a 2 + 6 2 _ 


6. a; + y - 

cd 


a 2 - b 2 

2%y 


x + y 


‘{-Z^b-aTb) 

8 




£ + 3 2 x -f 1 


11. If C = 

12. If C = 


E 


r + R 
E 

, R 
r + — 
n 


x — 2 2 x — 1 

, find C when E = f, r = f, and R = f. 

, find C when n = 40, E = 1.08, R = 500, 


and r = 4. 

13. By the formula F = i + - find F when w = 200 and 

rn w V 

v = 50. 


14. If b = —, and ^4 = 100, find 6 when : 

h 

a. h = 5; 6. A = 10; c. h = 20; d. h = 50. 

15. In the formula of example 14, if A has a fixed value, how 
does b change: 

a. When h increases ? b. When h is doubled ? 

16. If M = *L+A f find M when a = 4 and 5 = 2. 

ab 

17. What fractional part of 28 is 7 ? 6? w? 

Note. — Exercise 164, pages 286, 287, can be done now. 





















164 


ALGEBRA 


THE THREE SIGNS IN A FRACTION 

123. Three signs of any fraction must be considered, — 
the sign of the numerator, of the denominator, and of the 
fraction itself. Certain rules for making changes in these signs 
follow. 

Rule 1. — If the sign of either the numerator or the denom¬ 
inator of a fraction is changed, the sign of the fraction also 
must be changed. 


Thus: a. 


rjL=_l; but + A= + i 

+ 12 3 +12 3 


In this example, when — 4 is changed to + 4, the value of the frac¬ 
tion changes from — ^ to + i. 

6. but^A=+l 

+ 12 3 - 12 3 

In this example, when + 12 is changed to — 12, the value of the 
fraction changes from — i to + i. 

This rule is used in the following example. 


Example. Reduce to lowest terms 


12 + 2x - 2 


x* 


Solution. 1. The terms of the denominator, not being in the same 
order as those of the numerator, must be rearranged. 

x 2 — 9 x 2 — 9 


Then 


12 + 2 x — 2 x 2 — 2 x 2 + 2 x + 12 


2. The negative coefficient of — 2 x 2 is inconvenient. Change the 
signs of the denominator by multiplying by — 1. Then by Rule 1, 

x 2 — 9 x 2 — 9 


3. 


2 x 2 + 2 x + 12 2 x 2 - 2 x - 12 

_ _ (x - 3) (x + 3) 

2(x 2 — x — 6) 

1 

__ (x-3Kx + 3) s + 3 

2(x—-3-)(x + 2) 2(x + 2) 

1 


4 . 














FRACTIONS 


165 


Rule 2. — If the signs of both the numerator and the de¬ 
nominator of a fraction are changed, the sign of the fraction is 
not changed. 


Thus,-= — -; and also JlA 

+ 12 3 -12 


1 

3 


In this example, — 4 is changed to + 4, and + 12 to — 12, but the 
value of the fraction remains — + 


Similarly, —-— 

- 4 y 


2x 
4 y 


= - .5-; and +—— 
V - 2 1 


= + 3. 


124. There are two ways to change the signs of an expression 

(such as the numerator or the denominator of a fraction). 

a. If the expression is not factored, change the sign of every 
term by multiplying each of them by — 1 . 

Thus, (- 1)(- 2x 2 — 3z+4)=2z 2 +3z — 4. 

b. If the expression is factored, change the signs of all the terms 
in an odd number of factors of the expression. 

Thus: 2 X(- 3) X 4 = - 24, but 2 X(+ 3) X 4 = + 24. 

Similarly, consider (x — y)(y — z). 

y — z becomes — y z or z — y when its signs are changed. 

Therefore (x — y) (y — z) = — (x — y) (z — y), for changing the signs 
of y — z changes the sign of the product. 

On the other hand, (x — y)(y — z) = -\-(y — x)(z — y) because the 
signs of x — y and of y — z have been changed. This means that — 1 
has been used twice as a factor, and therefore the sign of the product 
is not changed. 

Do not confuse “ changing the signs of the terms of an ex¬ 
pression ” and “ rearranging the terms of an expression. ,, 

Thus, 2 x — 3 becomes 3 — 2 x when its signs are changed. 

2 x — 3 becomes — 3 + 2 x when its terms are rearranged. 

Note. — Changing the signs of the terms of an expression changes 
the value of the expression. Rearranging the terms of an expression 
does not change the value of the expression. 






166 


ALGEBRA 


EXERCISE 92 


Simplify the following fractions : 


1. 

— 3 x 
+ 4x 

4. 

, (+ 2)(— 3) 

- 12 

7. - 

2. 

1 

1 + 

t— 1 00 

5. 

_ (— 3 a) (+4 b) 
+ 24 ab 

8. - 

3. 

— 5 x 

6. 

{— 3 a)(— 4 6) 

9. - 

+ 20x 

- 24 ai 

10. 

Reduce to lowest terms ^- 3)(x -f- 2) 

(3 - x)(x - 2) 


(7s)(-3 y) 
+ 84 ary 

(— 12 m) 
(—15)(+4m) 
— 33 a& 
(-llo)(-6) 


Solution. Changing the sign of the factor (3 — x) of the denom¬ 
inator, makes it x — 3. But this changes the sign of the denominator 
and therefore of the fraction. 


. (s - 3)(s + 2) = _ (g-31(x + 2) _ x + 2 

(3 — x) (x — 2) (x—-3)(x - 2)’ x-2 

This is an inconvenient form for the result. If the signs of x — 2 be 
changed, then 

_ x + 2 — _i_ x —(— 2 
x-2 2-x 


. (x - 3)(x + 2) = _ (x - S)(x + 2) x + 2 

(3 — x)(x — 2) (g—-3)(x — 2) 2 -x’ 


Reduce the following fractions to lowest terms. 
Recall the rule given in § 110. 


x — 

• 5 

16. 

(a — b) (b — c) 

21. 

6 -|- m - m 2 

15 - 

3 x 

2 (c — b ) 

m 2 + m — 12 

2x- 

- 10 

17. 

(c - dy 

22. 

15 c - 10 d 

15 - 

3 x 

cd — c 2 

2 d? - cd -3 c 2 

a — 

2b 

18. 

0 - y) 2 

23. 

4 m — 3 mx 

4 b 2 - 

- a 2 

(; y - z) 3 

CO 

CO 

H 

1 

y 2 - 

- X 2 

19. 

T? 

1 

S' 

1 

e 

24. 

12- x- x 2 

4x- 

- 4 y 

(b — a)(c — b) 

x 2 - x - 20 

x 2 — 

f 

20. 

(x - y)(y - z) 

26. 

y 2 — 5 y + 6 

t y - 

x) 2 

to 

^1 

ta 

1 

9-y 2 


(Continued on p. 167.) 



































FRACTIONS 


167 


Find: 


26. 


27. 


28. 


29. 


30. 


(? + 2c + 1 1 ■ 

— c 



c 2 - 1 c + 2 


1 

— a a + 2 



a 

- 2 o - 1 



m 

2 + 2 m + 1 . 

m + 2 


m 2 - 1 

1 - 

m 


x 2 ~y 2 

y ~ 

- X 

X 1 

' + 2 xy + y 2 

X — 

2 y 

X 

1 

Sssl 

+ 

X 2 



y x 2 + 2 xy + y 2 


36. Find 


31. 


32. 


33. 


34. 


35. 


,2 _ 


4 x + 4 . 4 


6 a;+15 2 a;+ 5 


x 2 — 

1 .3 - X 

6 + x 

- x 2 x - 2 

1 — a 

_ a 2 — 3 a + 2 

2 — a 

a 2 — 1 

H 

\ 

CO 

1 

•1 


4 — x x 2 
13 x — 2 x 2 


7 x + 12 
15 . 2 a;- 


a — b 


2 b + a 


4 x 2 
2 


b 2 — a 2 b + a 
Solution. 1. In the second fraction, the denominator — a 2 + b 2 is 
inconvenient since the coefficient of a 2 is — 1. Therefore, change the 
signs of this denominator, and of the fraction. (Rule 1, page 164.) 


+ 


2 b cl 


a + 25 


b 2 — a 2 b + a a — b a 2 — b 2 
(Complete this solution.) 


a + b 


6 4- 

2 

43. 

2 

3 

4 c 

o-2 

2 — a 

3 + c 

3 - c 

c 2 - 9 

3 x . 

2 y 

44. 

* + 

X 

8 

a — b 

b — a 

z + 2^ 

2 - x 

x 2 - 4 

2 x , 

X 

45. 

2x 

j 1 . 

| 2 

x-2 + 

1 — X 

X 2 - y 2 

x + y 

y - x 

4 x , 

3 

46. 

2 a? , 

- a 4 

a 

x 2 - 1^ 

1 — X 

a 2 — l 

0+1 

1 — a 

2 

5 

47. 

a 

6 + 

2 ab 

x + 1 

1 - X 2 

a — p 

6-o 

a 2 - b 2 

6 b , 

3 

48. 

* 1 , 


s 

b 2 - 4 ^ 

2-b 

s 2 - 1^ 

^+1 

1 — 5 


Note. —Additional examples are to be found on page 286. 
















































X. FRACTIONAL EQUATIONS 


125. Changing equations which involve fractions to equations 
which do not involve fractions. 

Example. Solve the equation 2n \n — \ n = 27\. 
Solution. 1. 2ti + ^n-fn = ^- 

2. To obtain an equation which does not have in it any fractions, 
multiply by 6, the L. C. D. of the fractions. 


M 6 6(2 n + | n - f n) = 6 X A^. 

3. 12n+3w-4n = 165. 

4. C. T. 11 n = 165. 

5. Du n = 15. 

Check. Does 30 + 7\ - 10 - 27£? Yes. 


tfx § = 165 


2 


EXERCISE 93 

Solve and check the following equations : 
x x 9 




2 / + ! = 


3. 


10 
20 
3 3 

2 x _ x _ _9_ 
5 4 “ 10 

5t _ 1 = 2t 

2 3 3 

2_r = r , 5 

3 4 3 

x _ _9 , x 

4 10 5 


S_s _ s _ 2 

8" 3 _ 3 

8 c _ 62 _ c 

Y~J 3 


7. — 


9. 


10 . 


12 . 


14. 


16. 


4 w _ 2 w , 2 
"5 3~ *" 


5 m 

IT 


11 . - = - - 




26 _ 2 m 
~5 ~ 
x _ 1 
4 2 

3 r 2 r 


13. f 


— = 10 - 


- — = 7 


6 

3c 

5 


15. 3 d 


3 w 


5d 

6 

4 w 


= d , 5 
2 ' 9 


4 

- =- K) 

3 9 


168 



FRACTIONAL EQUATIONS 


169 


126. The process of changing the given equation to one 
which does not have any fractions is called clearing the equation 
of fractions. 

Example . Solve the equation 

3a—1 4a-5 _, 7 x + 5 

4 5 10 

Solution. 1. To eliminate (get rid of) the fractions, multiply by 20. 

2. M 20 iff • ~ f) _ jafr. (I f ~ 5 ) = 90 .4 -f ip-. (7 s + 5) 

4 0 Tff 

3. 5(3 a — 1) — 4(4 a — 5) = 80 + 2(7 x + 5). 

4. 15 a - 5 - 16 a + 20 = 80 + 14 a + 10. 

5. 15 — x = 90 + 14 x. 

6. /. -15x= 75. 

7. D_i5 x = — 5 . 

EXERCISE 94 


Solve and check the following equations : 


x x , 3 x _ 17 

9. 

r+ 16 _ 

r + 4 

"52 2 


6 

3 

2 5y _7y _1 

10. 

3t + 5 . 2t + 3 _ 0 

4 6 4 


8 

4 

3 z + 5 + 2 _z + 16 

11. 

4 x — 1 

_a - 1 _ 7 

3 4 


3 

2 12 

. w — 8 |Q w + 12 

12. 

6< + 3 

0 

10 

1 

CO 

-1- 

6 7 


7 

3 

B. c ~ 6 + 1 - 2 c + 4 

13. 

8 z — 1 

_z + l 11 

8 9 


3 

2 8 

m + 4 _ 13 m + 9 

14. 

7y + 6 

14 t/ + 1 

* 2 2 10 


3 

4 

a — 5 12 — a _ 1 

16. 

12 z + 2 

+ 4z + 1 _ 3 

' 4 6 2 


11 

2 

, + H _ 7 _ , _ 

16. 

5-8 

s + 5 _1 

8 2 


9 

2 2 




































170 


ALGEBRA 


17. Solve the equation 


3 m 


9 m 


12 


2 . 
3 m 


Solution. 1. The L. C. D. is 12 m. 

3 m / r. m /rt rj \ 4 

2. M 12m ( 3 m ~ 5 ) ~ J1 =^mr 

3. 3 m(3 m — 5) — m(9 m — 7) =8. 
(Complete this solution and check it.) 


2 

- 3 -m 


10 _ 

i 

9 loo 
II 

t—* 


23. 

3 - 2r 

6 r + 1 _ 20 

m 


r 

f 3r 3 

3 

+ J_ = 

13 

24. 

5^ + 6 

75 - 1 _ 1 

4 m 

5 m 

20 

25 

5 5 2 

5 

1 

ICO 

+ 

SI- 

= 0 

25. 

28 14 

- 2c 4 

2 x 

3 c 

5c 3 

8 _ 

4 9 


26. 

6a — 5 

_ 9 a + 5 25 

t 

5 1 5 


2 

3 18 a 

3 _ 

A = 13 


27. 

z — 8 , 

3 z — 6 _ 01 

y 

5 y 


2 z + 

3 z 2 


28. 


x — 2 x + 2 x z -4 

Solution. 1. The L. C. M. of the denominators is x 2 — 4. 
Multiply both members by x 2 — 4. 


x + 2 
(a£-’=='4) 




(* - 2 ) 

- (#-—4) 


1 

= —-A) 


(aH="2) 

.♦. 2(x + 2) - 5(x - 2) = 2. 
(Complete and check this solution.) 


(a£—=~4) 


Note 1. — As soon as possible, do Step 2 mentally. Always write 
Step 3, however, carefully inclosing the numerators in parentheses as 
is done in this solution. 

Note 2. — When solving fractional equations, the only root obtained 
may be a number which makes one or more denominators zero. Thus, 
in this example, if the root obtained is 2, the denominators of the first 
and third fractions become zero, and then these fractions do not have 
any meaning. (See § 67.) In such cases this root is only an apparent 
root and the “equation” is only apparently an equation, for a condi¬ 
tional equation is an equation only if there is a number which satisfies it. 





















FRACTIONAL EQUATIONS 


171 


29. 

15 -5 

39. 

t 

3 


X - 3 

3t + 4 

4< + 1 


30. 

4 3 

40. 

2 cl 4~ 3 

2 cl 4~ 9 


m — 1 m — 2 

cl 4~ 8 

3 cl -(- 4 


31. 

x - 5 _ x 4- 2 

41. 

3 

x 4- 12 

X 

x 4- 2 x — 5 

x 4- 3 

1 K 

CO 

1 

- 3 

32. 

c + 7 15 _ 0 

42. 

5x4-1 

3 _ x 


c — 5 c — 5 

CO 

1 

CN 

5 2 x - 

- 3 

33. 

x — 5 _ x — 2 

43. 

m 

2 

= 0 

x + 5 x + 2 

3 m 4- 4 

7m - 4 

34. 

12 x 2 + 3 x — 1 _ . 

3 x 2 + 2 x — 4 

44. 

rH 

II 

1 4- 

2 « 

1 

« 


35. 

Qt _ 8t j 

45. 

2 

6 

1 

t - 2 t - 1 + 

t +1 

2/4-1 Z 

- 3 

36. 

x_x + l 2 x 2 — 7 

9 3 9 x - 1 

46. 

6r + 35 

1 

00 

-i 

II 

& 


37. 

1 _ 3 x — 6 

47. 

4 x 

3 x 

1 

to 

« 

1 

1 

SI 

1 

to 

2x - 3 

1 

to 

H 

1 

00 

" 10 

38. 

2 . x _ 5 
x 2 ” 2 

48. 

c 

c - 1 _3 


c - 1 

c 2 



49. 

50. 

51. 

52. * 


CL -f- 2 
2 r 


+ 


a — 2 
r 


r - 4 r + 4 
5 x 2 + 6 3 


x 2 - 4 x — 2 
y 2 + 5 y - 30 j 


a 20 
= a 2 -4 
3 r 2 + 16 
r 2 — 16 
. = 5x4-3 
x + 2 

y = 2y 


2/ 2 — 2 t/ — 15 2/4-3 2 / — 5 


63. 2£l±3x + i_ = x + 1 
3 x 


54. 


2x4-1 
x _ x 2 4~ x — 4 
x — 2 x 2 — 4 


+ 


x 4- 2 






















































172 


ALGEBRA 


EXERCISE 95 

Solve and check the following equations: 
3 x — 2 _ 36 — 4 x 2 + 3 x 


x + 3 


x 2 — 9 


3 — x 


Suggestion. — In order to have all the expressions in descending powers 
of x, change the sign of the numerator of the second fraction, and of 
the denominator of the third fraction; also change the signs of these 
fractions. Then 

3 x — 2 4 x 


x + 3 


x 2 — 9 


36 _j_ 3 x + 2 


2 . 


3. 


4. 


V + 2 y - 2 


i - y 

4-y 2 


to 


1 


2 + to 
x + 5 
x — 3 
3 


4 — to 2 
4 


+ 


to + 3 


TO 


+ 


10 . 


ll. : 


12 . 


13 / 


14/ 


15. 


16/ 


3 — x 
4 

1 - 2r 
1 

TO + 1 


= 5 


1 


r + 4 
1 


6 . 


8 . 


9. 


x - 3 

\ 

+ 


1 


3 £ — 5 
3 TO 

2 to — 6 


2—2 2-3 

. 1 4 

' 2 


2 £ 


3 z — 4 
6z +5 


+ 


15 


4 x + 5 


5 TO 
3 


■ 6 z 4 — 3 a; 
2 3 # 


2 x 2 —2 x l-x 2 x 2 — l 


- = 1 — 
TO TOr 


2y 2 - 7y + 3 y 

-3 l-2y 

3 

2 

1 1 - 9 

1+3 A 

1 - 3 A 

+ 9+ 2 -l 

6 z 2 + 23 

1 _ o 

4 z 2 - 12 x + 9 3 

: - 2x ~ 

ci + 2 

2 a - 3 _ 

d 2 - 26 

a - 4 

a + 3 

12 + a — a 2 

2/ + 1 

S/ 2 + 7 

_ 2 y - 1 

to 

1 

00 

4y 2 - 9 

2 J/ + 3 6 — 4 j/ 

22 + 1 

9z + 17 . l-2z_ 0 


2 z — 16 48 + 2 z 


22 + 12 


Note. — Exercise 165, page 288, can be done now. 


















































FRACTIONAL EQUATIONS 


173 


EXERCISE 96 

1 . Two thirds of a certain number is 20 more than one 
fourth of the number. What is the number ? 

2 . What number is such that eight fifths of it, decreased by 
50, gives a remainder of fourteen fifteenths of it ? 

3. What number increased by one eighth of itself equals the 
sum of 30 and three sixteenths of itself? 

4. One half a certain number equals the excess of five 
sevenths of the number over 6 . What is the number ? 

5. What number exceeds by 13 the sum of one third itself, 
one fourth itself, and one fifth itself ? 

6 . There are three consecutive integers such that the sum 
of the second and third exceeds one fourth of the smallest by 17. 
What are these integers ? 

7. One sixth of a certain number is 10 less than one half the 
same number. What is the number ? 

8 . There is a certain number such that the sum of one half 
of it, one third of it, and one fourth of it is 3 more than the 
number itself. What is the number ? 

9 . Separate 62 into two parts such that five ninths of the 
greater shall be 7 more than one half of the smaller. 

10 . The sum of two numbers is 44. If the larger be divided 
by the smaller, the quotient equals f. What is the number ? 

11 . One ninth of the result obtained by adding 12 to a certain 
number is one half the result obtained by subtracting 9 from 
the number. What is the number ? 

12 . There are three consecutive integers such that the sum 
of the second and third is 21 more than one fifth of the smallest. 
What are these integers ? 

13. The perimeter of a triangle is 41 in. The second side is 
nine sevenths of the first side; the third side is one half the 
second side. How long is each side ? 


174 


ALGEBRA 


14. One sixth of “11 more than a certain number,” added to 
one third of “10 less than the same number,” gives 1. What 
is the number? 

15. What number added to both terms of the fraction £ 
changes that fraction into one whose value is f ? 

16. What number added to the terms of the fractions £ and 
bb makes the resulting fractions equal ? 

17. A’s age is twice B’s age. If A’s age 4 years from now be 
divided by B’s age at that time, the quotient is £. What are 
their ages ? 

18. A and B are carrying loads of 40 pounds and 50 pounds 
respectively. How many pounds must be taken from A’s load 
and added to B’s so that A will have three fifths as much to 
carry as B ? 

19. The denominator of a certain fraction is 5 more than the 
numerator. If the denominator be increased by 1, and the 
numerator be decreased by 4, the value of the resulting fraction 
is £. What is the given fraction ? 

20. The numerator of a certain fraction is 9 less than twice 
the denominator. If 1 be added to both terms of the fraction, 
the resulting fraction has the value What is the fraction ? 

21. There are two consecutive even integers such that the 
larger is f of the result obtained by subtracting 3 from the 
smaller. What are they? 

22. A’s age is two thirds of B’s age. Ten years ago it was 
one half of B’s age then. What are their ages now? 

23. There are three consecutive integers such that one fifth 
of the largest is one fourth of the result obtained by subtracting 
1 from the smallest. What are they ? 

24. There are three consecutive even integers such that one 
third of the smallest, plus one half of the second, plus one fourth 
of the third equals 15. What are they ? 


FRACTIONAL EQUATIONS 


175 


25. The denominator of a certain fraction exceeds its nu¬ 
merator by 22 . If 7 be subtracted from both terms of the 
fraction, the resulting fraction has the value What is the 
fraction ? 

26. A’s age is now two thirds of B’s age. A’s age 4 years 
from now will be five sixths of B’s present age. How old 
is each now ? 

27. A’s present age is one third of B’s. A’s age 10 years 
from now will be what B’s age was 10 years ago. How old is 
each now ? 

28. A’s age is now one fifth of B’s. In 12 years A’s age 
will be one half of B’s age then. How old is each now ? 

29. A man has $4.50 in nickels, dimes, and quarters. He 
has three fourths as many dimes as he has nickels, and five 
eighths as many quarters as he has nickels. How many coins 
of each kind has he ? 

30. The total value of a sum of money consisting of nickels, 
quarters, and half-dollars is $21.00. The number of half- 
dollars is two thirds of the number of nickels. The number of 
quarters is one half of the number of half-dollars. How many 
coins of each kind are there? 

31. A solution is made up of 45 pounds of water and 3 pounds 
of salt. How much salt must be added to make a 10% salt 
solution ? 

Suggestions 1. Let x equal the number of lb. of salt to be added. 

2. What is the total weight now? 

3. How much salt is there in the mixture? 

32. How much water must be evaporated from 40 pounds of 
a 4% salt solution to obtain a 5% salt solution ? 

33 . How much water must be added to one pint of a 50% acid 
solution to reduce it to a 40% solution ? 


176 


ALGEBRA 


127. Work problems. If a man can do a piece of work in 
8 days, then, in one day, he can do one eighth of it, and in 3 days 
he can do three eighths of it. 

If a man can do a piece of work in x days, then, in one day, 

1 15 . 

he can do - part of it, and in 5 days he can do 5 • - or - part of it. 


EXERCISE 97 

1. If a man can plow a field in 12 days, what part can he 
plow in 1 day ? in 4 days ? in n days ? in (x + 5) days ? 

2. If a man can do a piece of work in y days, what part can 
he do in 1 day ? in 3 days ? in z days ? 

3. A can do a piece of work in 6 days and B can do it in 
8 days. 

a. How much can A do in 1 day ? in # days ? 

h. How much can B do in 1 day ? in x days ? 

c. How much can they do together in 1 day ? in £ days ? 

4. A can do a piece of work in 8 hours and B can do it in 5 

hours. How long will it take them to do it if they work to¬ 
gether ? 


Solution. 

gether. 

2 . 


1. Let x = the no. of hr. it will take them to do it to- 


Since A does - in 1 hr., he will do - in x hr. 

8 8 

Since B does - in 1 hr., he will do - in x hr. 

5 5 

.*. together they will do - + - of it in x hr. 

8 5 

3. They complete the work in x hr.; that is, they do 4# or 1 times 
it in that time. 

••■!+!- L 


Note 1. 


(Complete the solution.) 
Observe especially Step 3. 


FRACTIONAL EQUATIONS 


177 


Note 2. — The essential facts of the solution can be charted. 



Person 

Time Alone 

Amount in 1 Hr. 

Amount in x Hr. 

A 

8 hr. 

l 

8 " 

X 

¥ 

B 

5 hr. 

1 

1 > 

X 

3 - 

Together 



tt.orl 



5. How long will it take A and B to do a piece of work 
together, if A can do it alone in 10 days and B in 20 days ? 

6. How long will it take A and B to do a piece of work 
together, if A alone can do it in 12 days and B in 9 days ? 

7. How long will it take A, B, and C to do a piece of work 
together, if A can do it alone in 10 days, B in 12 days, and C in 
15 days ? 

8. A man can plow a certain field with his team in 8 days. 
His son can do it with their tractor in 2 days. How long will 
it take them to do it together? 

9. A can do a piece of work alone in 10 days and B in 15 
days. After A had worked x days and B (x + 3) days, they 
found that they had done f of the work. How long had each 
worked ? 

10. In a factory, there is one machine which can turn out a 
certain number of articles in 8 hours, a second which can do it 
in 16 hours, and a third in 20 hours. How long will it take 
to fill a rush order for this number of articles, if all three ma¬ 
chines are worked at the same time ? 

11. A certain machine requires 25 hours to complete a certain 
job. After it had been running 17 hours, it became necessary to 
complete the work on a machine which requires 40 hours for the 
same job. How long should it have taken to complete the work ? 















178 


ALGEBRA 


128. Additional rate, time, and distance problems. 

EXERCISE 98 

1. After reviewing § 85, page 102, give a simple arithmetical 
example involving rate, time, and distance. 

2. What are the rule and the formula for finding 

a. the distance, when the rate and time are known ? 

b. the rate, when the distance and time are known? 

c. the time, when the distance and rate are known ? 

3. The rate of one train is f that of another. Let r repre¬ 
sent the rate of the slower train. 

a. Represent the rate of the faster train. 

b. Represent the time each requires to go 125 miles. 

4. Repeat Example 3 if the rate of the faster train exceeds 
that of the other train by 10 miles per hour. 

5. The rate of a passenger train is twice that of a freight 
train. It goes 200 miles in 2| hours less time than the freight 
train requires for a trip of 150 miles. Find the rate of each. 

Solution. Since the passenger train travels the more rapidly, its 
time for the trip will be less than that of the freight train. The prob¬ 
lem states it to be 2% hours less. 

Let r — the rate of the freight train. 


Train 

Rate 


Distance 

Time 

Freight 

r mi. per hr. 


150 mi. 

155 hr. 
r 

Passenger 

2 r mi. per hr. 


200 mi. 

?00 hr. 

2 r 


. 200 = 150 _ 2i 
2 r r 

(Complete and check this solution.) 











FRACTIONAL EQUATIONS 


179 


6. The rate of a certain passenger train is twice that of a 
freight train. The passenger train travels 140 miles in 2 hours 
less time than the freight train requires for a trip of 105 miles. 
What are their rates ? 

7. The rate of a certain passenger train is 10 miles more per 
hour than the rate of an automobile. It takes the train f as 
much time for a trip of 140 miles as it does the automobile. 
What is the rate of each ? 

8. The rate of one automobile party exceeds that of another 
party by 7 miles per hour. It goes 176 miles while the slower 
party is going 120 miles. What are their rates? 

9. The rate of a certain freight train is £ that of a passenger 
train. It takes the freight train lj hours longer than it does 
the passenger train to go 180 miles. Find their rates. 

10. The rate of a certain passenger train exceeds that of a 
freight train by 10 miles per hour. Its time for a trip of 210 
miles is £ that of the freight train for a trip of 100 miles. What 
are their rates ? 

11. The rate of a passenger train between two towns, M and 
N, exceeded the rate of an automobile by 13 miles per hour. 
The distance by railroad is 140 miles and by automobile, 165 
miles. If the time taken by the train is X 8 X of the time by 
automobile, find the rate of each. 

12. At 8 a.m. a train left C, traveling 36 miles per hour. A 
man who missed the train started in an automobile at 8 : 05 a.m. 
to overtake the train at D. If the distance to D by railroad is 
30 miles, and by auto 33 miles, how fast must the man travel to 
reach D at the same time as the train ? 

13. At 9 a.m. an automobile left P for Q, traveling 25 miles 
per hour. At 10 a.m. a second automobile left P for Q, travel¬ 
ing 30 miles per hour. a. If both reach Q at the same time, 
how far is P from Q ? b. At what time do they reach Q ? 


180 


ALGEBRA 


EXERCISE 99 

On a river, the direction in which the water is flowing is called down¬ 
stream, and the opposite direction is called upstream. When going 
downstream, a boat is carried along by the current of the river and 
whatever force is exerted within the boat; when coming upstream, 
its progress is retarded by the current of the river. 

1. The rate of the current of a river is 3 miles per hour. 

a. Express the rate downstream of some boys who are rowing 
at the rate of 5 miles per hour in still water? b. upstream? 

2. How long will it take the boys in Example 1 to go: 

a. 16 miles downstream? b. 16 miles upstream? 

c. 16 miles downstream, and back again? 

3. a. Express the time for the boys of Example 2 to go d 
miles downstream, b. d miles upstream. 

c. Write the equation expressing the fact that the total time 
for the trip is 5 hours. Solve the equation. 

4. Some boys who can travel 10 miles per hour in their motor 
boat want to make a trip on a river whose current is 3 miles per 
hour. How far may they go if they have only 7 hours for the 
round trip ? 

5. The rate of the current of a certain river is known to be 
3 miles per hour. Some boys in a motor boat found that it took 
them as long to go 8 miles downstream as it did to go 3 miles 
upstream. What was their rate in still water? 

6. How far downstream may a party go in a boat traveling 
12 miles per hour if the rate of the stream is 2\ miles per hour 
and if the round trip must be made in 4 hours ? 

7. A party went 20 miles downstream on a river whose 
current was known to be 2 miles per hour, in f- of the time it 
took them to return. What was their own rate in still water? 

8. Repeat Example 4 if the current of the river is 2f miles 
per hour. 


FRACTIONAL EQUATIONS 


181 


129. Literal equations. 

Example. 

1. M ; 


Solve - + - = c, for x. 


Solution. 

2. D c 


x x 
a + b = cx. 

a 4- b a 4- 6 

— 1 — = x, or x = —-L— 
c c 


EXERCISE 100 


Solve the following equations for x. 


1 . 

5 x = c 

5. 

r = px 

9. - = a 

12. 





X 


2 

2. 

ax = 2 

6. 

ax + b = 0 

10. * = b 

13. 

4 





a 


a; 

3. 

mx = n 

7. 

c — dx = 0 

.. m A 

11.- n — 0 

14. 

a 





X 


3 

4. 

rt = px 

8 . 

abx — c = 0 




15. 

Solve the equation ax — a 2 

= — b(b + x) for x. 



Solution. 

1 . 

ax — a 2 = 

— 6(5 -f- x). 



2 . 



a 

H 

1 

o- 

II 

- 6 2 - bx. 



3. 



ax + bx = 

a 2 - 6 2 . 



4. 



x{a + b) — 

(a + 6) (a — 6). 



5. 



.'. x = 

( “+W«- 6) ,ora 
(a + 6) 

- 6. 



a 

3 

3 

5 

b 

3 


Check this solution by substituting (a — 6) for x in the original 


equation. 

16. 7 x + 10 a = 16 x — 17 a 

17. 3 (x -2b) =83-11 b 

18. a(bx — a) = b(a — bx) 

19. rs( 1 — 2 x) = 2 s 2 x — r 2 

20. a(a — x) = b(b + x) 

21. a(x — a) = b(x — b) 

22. c(x — c + 2 d) = d(x + d ) 

23. r{x — s — r) = s(x —2 s) 

(Continued 


24. m?(x — 1) = mx — 1 

25. 2 ax + 6 b 2 = 3 bx + 4 ab 

26. 2 ax -f d = 3 c — bx 

27. b 2 = a(a + b 2 x — a 2 x) 

28. c(x — 2) = d(x — 1) 

29. a(x — a) = b(b — x) 

30. mx — m 2 = nx — n 2 

31. b 2 = a(a + bx — ax) 
l page 182.) 





182 


ALGEBRA 


32. — = 1 + - 


2 = 


33. 


12 a 2a: + 8a 


5 x 


-3 


38. 


4 a 


1+2 ax _ 2 a + 1 
2 a a 2 

= 1 


- 2 a 


34. —=!+'— 


39. 


35. 


x — 3 m _ 2 
3 


36. 


2 a: 

3 x — 5 a 


9 a; ■ 


2 x — 7 m 
’ 2~x 

7 « = _ 2 _ 
3a; 


12 


40. 


41. 


a: — 3 c 

a + 6_ 

2 

a: + a 


x — 5 c 


-2 
4 a 


x — 3 a x — 3 a 


= 5 


130. Equations involving decimals. 


Example. 

Solution. 

2 . 

3. 

4. D.m 


Solve the equation .7 a: — 1 = .36 x — .15. 


1. .7 x — 1 = .36 x — .15. 

.7 x - .36 x = 1 - .15. 

.*. .34 x = .85. 
x = 2.5. 


2.5 
?34)?85.00 

68 _ 

17 0 
17 0 


EXERCISE 101 

Solve the following equations : 

1. .5 x + .5 = 6.1 - .3 x 5. 2.1 - 1.5 2 = - .8 z - .7 

2. .9 m- .75 = .3 m - 3.75 6. 3 + 1.7 c = 2.7 c + 2.2 

3. .13 + .04 p = .11 p - .22 7. .3(8 - 5 a) = .2 (a - 5) 

4. .05 a - .12 = .16 - .09 a 8. .3(a; - 2) = 5(.2 x - 1.1) 

9. .5 = .3(a; - 2) - (a: - 6) 

10. 6(.2 - .3 a) + .3 = .3(4 a; - 5) 

11. .2 x - .03 x - .113 x = .01 + .161 

12. .3 x - .02 - .003 x = .7 - .06 x - .006 

13. .3(1.2 x - 5) = 14 + .05 a; 

14. .7(a: + .13) = .03(4 a: - .1) + .5 
.72 x - .55 


15. 3.3 a; 


.5 


= .1 x + 9.9 


Note. — Exercise 166, page 289, can be done now. 



















FRACTIONAL EQUATIONS 


183 


Ratio and Proportion 

131. Meaning and use of ratio. The ratio of one number to 
another is the quotient of the first divided by the second. 


The ratio of a to b is the fraction -; it is also indicated thus, 

b 

a : b. 


The first number (the antecedent) is made the numerator of 
the fraction, and the second (the consequent) the denominator. 

The ratio of two concrete quantities has meaning only if they 
are of the same kind. Their ratio is the quotient of their meas¬ 
ures in terms of the same unit of measure. 

Thus, the ratio of a segment 2 yd. long to a rectangle having 3 sq. 
yd. of surface is without meaning. 




EXERCISE 102 


What is the ratio of: 



1. 2 to 12 ? 

4. 

1 pt. to 2 qt. ? 

7. 3 d. to 2 wk. ? 

2. 3 to 15? 

6. 

3 oz. to 1 lb.? 

8. 6 in. to 3 ft. ? 

3. 2\ to 5? 

6. 

10 rd. to 1 mi. ? 

9. 1500 lb. to 1 T. ? 

10. Out of an 

income 

of $2200, a man 

spends $400 for rent. 


What is the ratio of his rent to his income ? 

11. A man’s total income was S3600 and his total expendi¬ 
tures were $3300. What was the ratio of his expenditures to his 


income ? 

12. A class consisted of 14 boys and 16 girls, a. What was 
the ratio of the number of boys to the number of girls ? 

b. What was the ratio of the number of boys to the total 
number of pupils ? c. of the number of girls ? 

13. How does the ratio of 15 to 18 compare with the ratio of 

20 to 24 ? with the ratio of 30 to 35 ? 

14. How does the ratio of 12 to 16 compare with the ratio of 

30 to 40 ? with the ratio of 33 to 44 ? 


184 


ALGEBRA 


132. Meaning and use of proportion. If the ratio of two 
numbers equals the ratio of two other numbers, the four num¬ 
bers form a proportion. 


Thus, | = Therefore 2, 4, 6, and 12 form a proportion. 

As a proportion , we read it, 2 is to 4 as 6 is to 12 ; meaning 2 has the 
same relation to 4 as 6 has to 12. 

Similarly, the numbers a, b, c, and d are proportional if y =-• 

b d 

Read it, a is to b as c is to d. It is also written a : b = c: d. 

a , b, c, and d are called the first, second, third, and fourth 
terms of the proportion, a and d are the extremes of the pro¬ 
portion ; b and c are the means of the proportion. 


Example. 

Solution. 

2. M n px 

3. 

4. Dm 


Find x from the proportion m:np = c : 
m _ _c_' 
np nx 

x v 

m 1 c 

-npxr- — =-npxr -- 

-np- nbx~ 

mx = cp. 

_ cp 


EXERCISE 103 


Find x in each of the following proportions: 


X _ 

5 

6 . 

X 

_ b 

11 . 

x — 1 

_ 5 

16. 

X 


m 

4 “ 

3 

a 

c 

2 

3 

X + 

a 

n 

2 a 

_ 3 

7. 

m 

_ n 

12 . 

x — 3 

_ 4 

17. 

X 


c 

X 

5 

X 

r 

x -b 5 

6 

a — 

X 

d 

a _ 

X 

8 . 

V , 

_ r 

13. 

X 

_ 3 

18. 

X — 

a 

c 

4 ” 

3 

q 

X 

to 

1 

a 

4 

b 


d 

5 = 

4 

9. 

r 2 

_ r 

14. 

X 

_ a 

19. 

x — 

a _ 

3 

c 

X 

sx 

~ t 

5 — x 

~ 2 

X + 

a 

4 

a _ 

2 

10 . 

ac 

_ c 2 

15. 

X 

_ 5 

20 . 

x — 

m 

a 

X 

b 

X 

~ d 

x — a 

4 

x + m 

b 












FRACTIONAL EQUATIONS 


185 


21. Eight workmen made 1432 articles in one day. How 
many men would be needed to make 2000 such articles at the 
same rate ? 

Suggestion. — If x is the number of workmen, then 
x = 2000 
8 1432’ 

Observe that x, the number of workmen needed for 2000 articles, 
is to 8, the number who made 1432 articles, as 2000 articles is to 1432 
articles. Complete this solution. 

22. If a family spent $18.75 for electricity in 5 months, what 
would be their expenditure in 12 months at the same rate ? 

23. A boy earned $31.50 in three weeks. At the same rate, 
what would he earn in 17 weeks ? 

24. The taxes on a piece of property valued at $6500 were 
$73.25. At the same rate, what would be the taxes on a piece 
of property valued at $8300? 

25. The net profits in a store which had sold $65,000 worth 
of goods were $4250. At the same rate, what would be the 
profits from the sale of $85,000 worth of goods ? 

26. When coal was selling for $9 per ton, one family spent 
$225 for fuel. What would the same amount of fuel cost at 
$15 per ton? 

27. Separate a line 45 inches long into two parts which have 
the ratio 4 to 9. (Solve it in the two ways suggested below.) 

Solution a. Let x = shorter part; and .*. 45 — x = longer part. 

Solution h. Let 4 x] = shorter part; and 9 x = longer part. 

28. Separate a line 90 inches long into three parts proportional 
to 2, 3, and 4. (The method used in Solution b above should 
be used in this case.) 

29. Separate 280 into four parts proportional to 2, 3, 4, and 5. 

30. Divide $2500 among three children so that their portions 
shall be proportional to 3, 3, and 4. 


186 


ALGEBRA 


133. Deriving and using formulas. 

Example. Derive a formula for the rate in terms of the 
amount, the principal, and the time in simple interest problems. 

Solution. 1. The formula for the amount is A = P + PRT. 

Solve this equation for R. 

2. S P A - P = PRT, or PRT = A - P. 


3. D pt 


R = 


A - P 
PT 


This formula enables us to find R when A, P, and T are known. Thus, 
if a man receives $3500 at the end of 6 years from an investment of 
$2400, what rate of simple interest has his money earned? 

Here, A = $3500, P = $2400, and T = 6. 


.*. R = 


3500 


2400 X 6 


2400 Qr 1100 


14400 


.*. R = .076+ or 7.6+%. 


EXERCISE 104 

Solve each of the following equations for the letters indicated: 

1. A = ab. a. Solve for a. b. Find a, if A = 96 and b = 15. 

2. A — — • a. Solve for b. b. Find b, if A = 420 and a = 24. 

2 

3. C = 2 7r r. Solve for r. 4. V = Iwh. Solve for w. 

5. V = i bh. a. Solve for b. b. Find b, if V = 475 and 
h = 39. 

6. F = f C + 32. a. Solve for C. b. Find C, if F = 77. 

7. A = — is a formula from geometry. 

a. Solve the formula for a ; b. for b ; c. for c. 

d. Find c when A = 550, b = 30, and a = 22. 

8. A = P + PR T. a. Solve for P; b. for T. 

c. Find T when A = $3500, P = $3000, and R = 5%. 

9 . T = - + tis a formula from physics, a. Solve it for a. 

b. Find a when T = 245.5 and t = — 27.5. 






FRACTIONAL EQUATIONS 


187 


10. mg — T = mf is a formula from physics. 
a. Solve it for T ; b. for /; c. for m. d. Using the result 
of part c, find m when g = 32.16, T = 7040, and / = 4. 

( Tfi \ 

——— )v. a. Solve it for m. b. Find m when 
M + m/ 

u = 35, M = 32, and v = 750. 

12. C = ~- a ^ ■ • a. Solve it for a. 
b — a 

b. Find a when C = 13,680, K = 270, and b = 9.5. 
c — b 

13. 5 = — —- is a formula from physics. 
a — b 

a. Solve it for b; b. for a. c. Using the result of part a, 
find b when s = 2.7, c = 45, and a = 26.3. 


14. h = &(1 + -g-ig- 1) is a formula from physics. 
a. Solve it for t. b. Solve it for h. 


15. I = 


M 


Mt 


is a formula from physics. 


a. Solve it for L ; b. for M. c. Using the result of part a, 
find L when l = .000012, M = 100, and t = 75. 


1 c 3000 ad 2 » t p • 

16. w = - ---- is a formula from engineering. 

600 d 2 + l 2 5 6 

a. Find w , correct to two decimals, when a = 18, l = 12, 

and d = 10. b. Solve the formula for a. 


17. p = 


ad 2 

~T 


+ d is a formula from engineering. 


a. Find p when a = .56, d = -J, t = -f. 

b. Solve the formula for t. 

18. C = —^— is a formula from physics. 

nR + r 

Solve it for n. 

19. F = is a formula from physics. 

gr 

Find F when m = 150, v = 25, g = 32, and r — 5. 








XI. GRAPHICAL REPRESENTATION OF STATISTICS 
AND OF MATHEMATICAL FUNCTIONS 


134. Review of graphs taught previously. 

The figures below recall bar and broken line graphs. 


a. Vertical Bar graph 


b. Horizontal Bar graph 


Graph showing the number 
of million dollars of sales of 
chain stores. 


Graph showing the number 
of million dollars of sales of 
mail order houses. 




c. Broken Line graph 
The graph at the right ap¬ 
peared in a monthly magazine 
for business men. Observe 
that it gives the cash price for 
wheat, and also for corn, for 
each year from 1918 to 1926, 
and for each month of 1927 
from January to September 
inclusive. 



Chicago grain prices 

-3.00 

J2-50 

1,2.00 

*1.50 

“1.00 

.50 





Cash wheat- 

Cash corn -- 






d 












- 3.00 
-2.501 


r 

1 





L 









jj 

n 


/ 








- 2.00 p, 

1 50 | 




\ 

b 

Jj 















’’ >/, 








1918 ’20 ’22 ’24 ’26jfmamjjasond 


188 

















































































































































































































GRAPHICAL REPRESENTATION 


189 


EXERCISE 105 


1. Draw a horizontal bar graph of the statistics in the table 
below : 


In the Yeab 

1920 

1921 

1922 

1923 

1924 

1925 

1926 

The no. of lb. of coal 
used to make one K. 
W. H. of electricity was 

5 

4.6 

3.8 

3.5 

2.6 

2.4 

2.2 


2. Draw a vertical bar graph of the statistics in the table below : 


In the Year 

1920 

1921 

1922 

1923 

1924 

1925 

1926 

The no. of million 
K. W. H. of electricity 
sold by one company 
was. 

350 

400 

520 

650 

750 

900 

1550 


3. Draw a vertical bar graph to represent the monthly 
savings of a family for one year. 


Jan. . . 

. $11.50 

May . . 

. $8.25 

Sept. 

. $15.00 

Feb. . . 

7.50 

June . . 

. 4.25 

Oct. . . 

6.75 

Mar.. . 

8.00 

July . . 

. 9.00 

Nov.. . 

9.25 

Apr. . . 

6.00 

Aug. . . 

. 5.50 

Dec. . . 

. 10.00 


4. The normal weights of boys and girls of certain ages : 


Age 

6 Yr. 

7 Yr. 

8 Yr. 

9 Yr. 

10 Yr. 

11 Yr. 

12 Yr. 

13 Yr. 

14 Yr. 

15 Yr. 

Boys 

Girls 

45 lb. 
43 lb. 

49 lb. 
47 lb. 

54 lb. 
52 lb. 

59 lb. 
57 lb. 

65 lb. 
62 lb. 

70 lb. 
69 lb. 

76 lb. 
78 lb. 

84 lb. 
88 lb. 

95 lb. 
98 lb. 

107 lb. 
106 lb. 


On the same set of axes, draw two graphs, one representing 
the weights of boys, and the other the weights of girls. 


5. Represent these temperatures at New York City in a 
recent year, by three graphs drawn on the same set of axes. 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

Temperature 

Highest 

67° 

69° 

78° 

91° 

95° 

97° 

99° 

102° 

100° 

88° 

74° 

68° 

Lowest 

-6° 

-7° 

3° 

20° 

34° 

45° 

50° 

51° 

36° 

31° 

7° 

-3° 

Average 

30° 

31° 

38° 

48° 

59° 

68° 

74° 

72° 

66° 

56° 

44° 

34° 



































































190 


ALGEBRA 


135. A distribution graph shows the relation of parts to a 
whole and to each other. Two most common distribution 
graphs are shown below. 

Example 1. Circular graph showing the distribution of our 
population in a recent year. 

Explanation of this graph. — The 
total population of the United 
States is represented by the interior 
of the circle. The foreign-born 
whites, 23% of the population, are 
represented by 23% of the interior 
of this circle; etc. 

How this graph was made .— 

1. First a table of statistics gave 
the number of each part of the population. From this table, 
it was necessary to determine, for example, that the foreign- 
born whites were 23% of the total population. 

2. Next, 23% of 360° was computed. This is about 83°. 

3. An angle of 83° was made at the center of the circle, by 
means of a protractor. The part of the interior of the circle, 
inside this angle, then represents the foreign-born white part 
of the population. Etc. 

Example 2. Rectangular graph showing the distribution of 
our population, using the same statistics. 

Explanation of graph. — The various parts of the population 
are represented by parts of the interior of the rectangle. 

Native Whites Native Whites Foreign-born 

Native White Parents Foreign Parents Whites Negroes 


47.5% 26.6% 23% 

Construction. — The rectangle was made 3J inches long; 47.5% 
of it, or 27 sixteenths of an inch , was computed. A part that long was 
used to represent the native whites of native white parentage, etc. 







































GRAPHICAL REPRESENTATION 


191 


EXERCISE 106 

Represent either by a circular graph or by a rectangular graph 
the following sets of statistics : 

1. In a certain class, of the pupils were marked Excellent; 
i, Good; f, Fair; -J-, Poor; and the rest, failed. 

2. In 1920, the United States furnished about 68% of the 
raw cotton for the world; British India, 13%; Egypt, 7%; 
China, 5%; all other countries, 7%. 

3. In a recent year, the United States consumed about 29% 
of the world’s supply of cotton; Great Britain, 12%; India, 
11%; European countries, 28%; Japan, 11%; all other coun¬ 
tries, 9%. 


4. In 1920, of the population of the United States : 



11% 

21% 

18% 

29% 

16% 

5% 

was 

under 5 yr. 

5 yr. 
to 

14 yr. 

15 yr. 
to 

24 yr. 

25 yr. 
to 

44 yr. 

45 yr. 
to 

64 yr. 

over 65 yr. 


5. In a certain school, four teachers gave to their pupils the 
marks indicated in the table following : 


Teacher 

Ex. 

Good 

Fair 

Poor 

Failed 

Total 

F 

3 

? 

4 

? 

3 

? 

5 

? 

3 

? 

? 

K 

2 

? 

4 

? 

5 

? 

1 

? 

2 

? 

? 

M 

3 

? 

6 

? 

4 

? 

2 

? 

2 

? 

? 

W 

4 

? 

4 

? 

9 

? 

3 

? 

4 

? 

? 


a. Find the total number of pupils in each class. 

b. How many per cent of the pupils in each class received 
each grade? Insert where the question marks appear. (Ex¬ 
press these per cents to the nearest per cent; thus, call 16.3% 
16%.) 

c. Make a separate distribution graph for each teacher. 




























192 


ALGEBRA 


EXERCISE 107 

Optional Review of Statistical Graphs 

1. Below is given a table which shows the weekly wage at 
various ages, from 14 years to 25 years, of boys who leave school 
at the age of 14 and boys who leave school at the age of 18: 


Weekly Wage of a Boy Weekly Wage of a Boy 


Age in Who Leaves School Who Leaves School 

Years at the Age of 14 at the Age of 18 

14 $4.00 

16 5.00 

18 7.00.$10.00 

20 9.00 15.00 

22 . 11.00 . 20.00 

24 12.00 24.00 

25 13.00 30.00 


On the same diagram make a graph of the two sets of data, 
using a broken line for each. 

2 . The average weight of boys at different ages from 6 years 
to 15 years is given in the table below: 

Age. 6 7 8 9 10 11 12 13 14 15 

Weight (in pounds) . 50 53 57 62 67 72 78 85 93 105 

a. Make a broken line graph of this table. 

b. A boy, normal in every respect, weighs 87 pounds ; what is 
his approximate age ? Show how this is found from the graph. 

3. The temperature readings for a certain town from 8 a.m. 
to 6 p.m. for one day are given in the following table: 

Hour . . 8 9 10 11 12 1 2 3 4 5 6 

Reading . 10° 12° 16° 23° 32° 40° 43° 43° 41° 34° 29° 

a. Make a broken line graph of the above table. 

Determine from the graph: 

b. The approximate temperature at 11.30 a.m 

c. The time when the reading was 35°. 

d. During what hour there was no change in temperature. 














GRAPHICAL REPRESENTATION 193 

4. In a certain high school the number of pupils in the four 
classes was distributed as follows: 

First year 690 pupils Third year 420 pupils 

Second year 750 pupils Fourth year 300 pupils 

Represent this information by the use of a rectangular or circular 
distribution graph. 


5. The following table shows the earnings and spendings of a 
boy from April to October inclusive during a certain year : 


Month 

April 

May 

June 

July 

August 

Sept. 

Oct. 

Earnings .... 

$2.00 

$3.20 

$4.80 

$10.40 

$12.00 

$4.40 

$5.20 

Spendings .... 

3.20 

3.60 

2.40 

2.00 

1.60 

8.40 

4.00 


a. On the same diagram, make a full broken-line graph to repre¬ 
sent earnings and a dotted broken-line graph to represent spendings. 

b. Mark on the drawing one place where the two graphs cross. 
Read this point and tell what it means. 

6. A grocer’s record of sales of peaches for a certain year 
showed that at $1.25 a basket he sold 10 baskets; at $1, he sold 
30 baskets; at 75^, 70 baskets; at 60^, 85 baskets; at 50j£, 90 
baskets. Represent this graphically, and from your graph 
estimate the probable demand when the price is $1.10 a basket. 


7. Represent graphically the strengths of various woods : 


Wood 

Poplar 

Yellow 

Pine 

White 

Pine 

Red 

Fir 

Walnut 

White 

Oak 

Maple 

No. OF LB. TO 

EACH SQ. IN. 

7,000 

11,000 

15,000 

10,000 

12,000 

14,500 

10,500 


Suggestion. — Let i in. = 1000 lb., or let 4 spaces = 1000 lb. 

8. Represent the distances through which an object, falling 
freely, falls in various times : 


In. 

0 sec. 

1 sec. 

2 sec. 

3 sec. 

4 sec. 

5 sec. 

It falls 

Oft. 

16 ft. 

64 ft. 

144 ft. 

256 ft. 

400 ft. 









































194 


ALGEBRA 


GRAPHS OF MATHEMATICAL RELATIONS 
136. Definitions. In the figure below: 



a. lines XX' (read ex ex prime) and YY', drawn at right 
angles, are called the axes. 

b. XX' is called the horizontal axis or A-axis. 

c. Y Y' is called the vertical axis or 7-axis. 

d. Point 0, where the axes intersect, is called the origin. 

e. PR, perpendicular from P to XX', is the ordinate of P. 

f. PS, perpendicular from P to Y Y', is the abscissa of P 
(plural, abscissas). 

g. The ordinate (PR) and abscissa (PS), together, are the co¬ 
ordinates of P. 

Since ORPS is a rectangle, PR = OS; therefore the ordinate 
of P equals OS. Similarly, the abscissa of P = OR. 

Distances up from XX' are positive; distances down from 
XX' are negative. Notice on OY the positive scale, + 1 
+ 2, + 3, etc., and on OY' the negative scale, — 1, — 2, etc. 



































































































GRAPHICAL REPRESENTATION 


195 


Distances to the right from YY' are positive ; distances to 
the left are negative. Again notice the positive and negative 
scales on XX'. 

Thus, the abscissa of P is + 3; and the ordinate of P is + 4. P 
is called the point (+3, +4). This means that P is located 3 spaces 
to the right and 4 spaces up. 

When giving the coordinates of a point, the abscissa is always 
given first. 

The scale need not be the same on the two axes. 

Note. — The numbers of the scale should always be directly below 
the points of the horizontal axis, and exactly to the left of the points 
of the vertical axis. 

EXERCISE 108 

1. What is the abscissa of H ? F ? P ? H ? 

2. What is the ordinate of F ? P ? P ? 27? 

3. What are the coordinates of H? F? P? P? H ? 

4. On a similar sheet of paper, or on the blackboard, locate 
the following points: 

a. (3, 5) b. (— 4, 5) c. (— 3, — 6) d. (+ 4, — 3) 

e. (0, 7) /. (+ 5, 0) g. (0, - 10) h. (-8, 0) 

6. a. Locate points R : (2, 3) and S: (— 5, — 3). Draw RS. 

b. Locate points T: (— 3|, lj) and W : (3, — 1). 

c. What are the coordinates of the point where RS intersects 

TW? 

6. a. Locate C : (5, 4); D : (— 5, + 6); E : (— 7J, — |); 

F: ( 2 ^, — 2 ^). 

b. Draw CD, DE, EF, and CF. What figure is formed? 

c. Draw its diagonals. What are the coordinates of their 
point of intersection? 

7. a. Locate M : (+ 6, - 4) and N: (-6, + 2), and draw MN. 

b. What are the coordinates of the mid-point of MN? 

8. Repeat Example 7 if M is (+ 5, +4) and N is (— 6, — 3). 


196 


ALGEBRA 


137. Variables. In each of the examples of Exercises 105- 
107, there was a table of values which some changing number 
had under certain changing conditions. 

A changing number is called a variable. The variable is 
said to vary with or vary according to changing conditions. 

138. If the values of one variable depend upon the values of 
one or more other variables, then the first variable is said to be a 
function of the other variable or variables. 

139. Graphing the relation between two variables connected 
by a formula or equation. 

Example. Represent graphically the relation between x 
and y, when x and y satisfy the equation 2 x + y = 4. 

Solution. 1. Select values of x and determine the corresponding 
values of y. 

Thus, when x = — 3, 2(— 3)+ y = 4; — 6 + y = 4; y = 10. 
when x = + 6, 2(6) y = 4; 12 + y = 4; y = — 8. . 

Similarly, 


When x = 

- 3 

- 1 

0 

+ 2 

+ 4 

+ 6 

Then y = 

+ 10 

+ 6 

+ 4 

0 

- 4 

- 8 


2. Use each pair of values so obtained as coordinates of a point. 

Thus, place a dot at (+6, — 8) ; at (— 3, + 10); etc. 

3. Draw the smooth line connecting these points. 



4. This is the required graph. Notice that it is a straight line. 















































































GRAPHICAL REPRESENTATION 


197 


EXERCISE 109 

1. a. What are the coordinates of C? Do they satisfy the 
equation ? 

h. Similarly, test the coordinates of B and D. 

2. a. Consider point (5, 5), not on the graph. Do these co¬ 
ordinates satisfy the equation ? 

b. Try the coordinates of any other point not on the graph. 

3. a. Express graphically the relation y = x — 3. 

Hint. — Let x = — 3, — 2, — 1, 0, 1, 2, 3, 4, 5. 

b. What does the graph appear to be? 

c. Select a point not on this graph. Do its coordinates 
satisfy the given equation ? 

4. a. Express graphically the relation x + 2 y = 6. 

Hint. — Follow the steps of the illustrative example. Let x — — 4, 
- 2, 0, 2, 4, 6, 8, 10. 

b. What does the graph appear to be ? 

c. Select some point on the graph. Do its coordinates satisfy 
the given equation ? 

d. Select a point not on the graph. Do its coordinates satisfy 
the equation? 

6. Repeat Example 4 for the equation x — 2 y = 6. 

6. Draw the graph of the equation x + y = 5. 

7. Draw the graph of the equation x — y = 8. 

140. Summary, a. The graph of an equation of the first 
degree having two variables is a straight line. 

For this reason these equations are called linear equations. 

b. The coordinates of every point on the line satisfy the 
equation. 

c. The coordinates of every point not on the line do not 
satisfy the equation. 


198 


ALGEBRA 


141. Graphs drawn in an engineering course. 



















GRAPHICAL REPRESENTATION 


199 























200 


ALGEBRA 


EXERCISE 110 

Review of Graphical Representation of Formulas 

1. a. Write the formula expressing the interest, 7, on $100 at 
5% for n years. 

h. Make a table of values of I when n = 1, 2, 3, ... 10. Draw 
the graph, using the horizontal axis for the values of n. 

c. From your graph, determine the interest on $100 at 5% 
when n = 3^- yr.; when n = yr.; when n — 6J yr. 

2. a. Write the formula expressing the selling price, S, of an 
article which cost C dollars, if the selling price is to equal one 
dollar more than f of the cost. 

b. Make a table showing the values of S when C = 0, 4, 8, 
12, 16, and 20. Represent this relation graphically, using the 
horizontal axis for the values of C. 

c. From your graph, find S when C = 6; when C = 9; 
when C = 17. 

d. How does S change when C increases ? 

e. How much does S increase when C increases $1.00? 

3. a. Write the formula expressing the total wages, W, of a 
workman who works n hours at 85jzf per hour. 

b. Represent this relation graphically, for n = 0, 8, 16, 24, 
32, 40, 44. 

c. How does W increase when n increases one unit? (This 
means, increases from, say, 3 to 4.) 

d. How does W change when n is doubled ? Trebled ? 

4. The formula for the area, A, of a square whose side is s 
is A = s 2 . 

a. Make a table of values of A for s = 0, 2, 4, 6, 8, and 10. 

b. Draw the graph, using the horizontal axis for the values of s. 

c. From the graph, determine A when s = ,5; when s = 7.5. 

d. From the graph, determine s when A = 50; when A = 90. 

e. How does A change when s is doubled ? 


XII. LINEAR EQUATIONS HAVING TWO UNKNOWNS 

142. A linear equation or first degree equation having two 
or more unknowns is one in which no unknown appears in the 
denominator of a fraction, and in which the sum of the ex¬ 
ponents of the unknowns in each term is only 1. 

Thus, x + y — 10 is a linear or first degree equation. 

But xy + x = 10 is not a first degree equation since the sum of the 
exponents of x and y in the first term is 1 + 1 or 2. 

Also ^ + y = 10 is not a first degree equation in x and y because x 

appears in a denominator. When the equation is cleared of fractions, 
1 + xy = 10 x, and this again is of degree two. 

143. A solution of an equation having two unknowns is a set 
of values of the unknowns which together satisfy the equation. 


Thus, in the equation x + y = 10, if x = 3, then 3 + y = 10, and 
therefore y = 7. x = 3, y = 7 is a solution of x + y = 10. 

Similarly, in the equation x + y = 10 


If X * 

1 

2 

4 

7 

10 

15 

- 2 

etc. 

then y = 

9 

8 

6 

3 

0 

- 5 

+ 12 

etc. 


Each of these pairs of numbers is a solution of x + y = 10. 

An equation with two unknowns has an infinite number of 
solutions. For each value of one unknown, there is a cor¬ 
responding value of the other unknown. As one unknown varies 
(changes in value), the other also varies. The unknowns, there¬ 
fore, are called variables (changing numbers). 

Therefore these equations are called, preferably, equations 
having two variables. 


201 
















202 


ALGEBRA 


144. Short method of drawing the graph of a linear equation 
having two variables. In § 140, you learned that the graph is 
a straight line. Since a straight line can be drawn with a 
ruler as soon as two of its points are known: 

Rule. — To draw the graph of a linear equation having two 
variables: 

1. Select a value for one variable. Determine the corre¬ 
sponding value of the other variable. These are the coordinates 
of one point on the graph. Locate it. 

2. Locate a second point in the same manner. 

3. Draw the straight line passing through these two points. 

4. Find a third point as in Step 1. If it falls on the graph, 
the work has been done correctly. 

Example. Draw the graph of 4 x — 3 y = 6. 

Solution. 1. Let x = 0. Then y = — 2. J Obtained mentally. 

2. Let x = 3. Then y = 2. \ See Note, page 203. 

3. Locate the points and draw the graph. 



Check . Let x = 6. Then y = 6. Does this point lie on the line? Yes. 
(Read the Note on page 203.) 
























































































EQUATIONS HAVING TWO UNKNOWNS 203 


Note. — The easiest value of x to use is x = 0, as in this example, 
for then —3^ = 6 and y = — 2. 

In an equation like 3 x — 5 y = 11, when x = 0, —5 y = 11, and 
y = — While this is correct, the point (0, — 2|) cannot be located 
easily. In such cases, try y = 0. Then 3 x — 11, and x = 3f. This 
is equally unsatisfactory. 

Try x = 1, then 3 — 5 y = 11; — 5y = 8; y = — If. Unsatis¬ 
factory. 

Try x = 2, then 6 — 5 y = 11; — 5y = 5; y = — 1. This is a 
satisfactory point. 

Try x = — 3, then — 9 — 5 y = 11; — 5y = 20; y = — 4. 

Or, from 4 x — 3 ?/ = 6, y = %x — 2. Let x = 3, 6, — 3, etc. Then 
?/ will be an integer. 

If necessary, of course, use fractional values of x or y, or both. 

EXERCISE 111 

Draw the graph of each of the following equations, each on a 
separate piece of coordinate paper 


1. X — y = 5 

4. \ x — y = 5 

7. 4 x + y = 0 

2. 2 x - y = 3 

5. 2 y = 6 — x 

8. 4 x + 5 y — 20 

3. 3 x + y = 4 

o 

II 

CO 

1 

>> 

CO 

9. 3 x - 2 = y 

10. 

3 M + 5 N = 15 


Hint. — Solve as if M were x and N were y. 

Place values of M along 

Drizontal axis and those of N on vertical axis. 


11. y = x + 2.5 

13. s = 1.25 c 

15. C = 3.1 D 


12. y — .75 x 14. y — 5 — .75 x 16. s — 1.5 c + 5 

17. a. The sum of two numbers is 9. Letting x and y repre¬ 
sent these numbers, form the equation expressing the fact that 
their sum is 9. Then draw the graph of the equation. 

b. From your graph find y when x is 3.5. 

18. a. One number exceeds twice another number by 5. Ex¬ 
press these numbers by x and y; form the equation; draw its 
graph. 

b. From your graph, find the larger when the smaller is 6. 

c. Find the smaller when the larger is 11.5. 


204 


ALGEBRA 


145. Two linear equations having two variables. 

Example. Draw upon the same set of axes the graphs of 
f 2 x — y = 4 
1 2 z + 3 y = 12 


Solution. 1. Draw the graphs of the two equations. 


(1) 2 x — y — 4 

(2) 2 x + 3 y = 12 

Point A 

When x = 0, y = — 4 

Point D 

When x = 0, y = 4 

Point B 

When x = 2, y = 0 

Point E 

When x — 6, y = 0 

Point C 

When x = 4, y = 4 

Point F 

When x = — 3, y = 6 



2. Recall § 140 6, page 197. 

Observe point G. Its coordinates are x = 3, y = 2. 

(3, 2) satisfy 2 rc — y = 4, since 2-3—2 does equal 4. 

(3, 2) satisfy 2x + 3 y — 12, since 2 • 3 + 3 • 2 does equal 12. 

.*. the coordinates of G satisfy both equations. 

3. Since no other point lies on both lines, (3, 2) is the only pair of 
numbers which will satisfy both equations. 

4. These equations are called independent equations because each 
has solutions which are not solutions of the other; they are also simul¬ 
taneous equations (a special kind of independent equations) because 
they do have one common solution. 








































































































EQUATIONS HAVING TWO UNKNOWNS 205 


146. Two equations having two variables are independent 


if each has solutions which are not solutions of the other. 


They are said to form a system of independent equations. 

Two independent equations which do have common solutions 
are called simultaneous equations. 

To solve a system of simultaneous equations is to find their 
common solution or solutions. 

147. Two simultaneous linear equations having two variables 
have only one common solution. 

Rule. — To determine graphically the common solution of 
two simultaneous linear equations having two variables : 

1. Draw the graphs of the two equations upon one set of axes. 

2. Determine the coordinates of the point which is on both 
graphs. This is the common solution. Check the common 
solution by substituting it in both equations. 


EXERCISE 112 

Determine graphically the common solution of: 



Note. — Sometimes it is difficult to find the common solution accu¬ 
rately by this graphical method. Example 12 may have caused you 
some difficulty. Better ways follow in § 148 and § 149. 


206 


ALGEBRA 


148. Elimination by addition or subtraction. 

Example 1. Solve the system j 7 ^ + 4 y — ^2 


Solution. 1. M 4 * (1) 20 x - 12 y = 76. (3) 

2. M 3 (2) 21 x + 12 y = 6. (4) 

3. Add (3) and (4) 41 re = 82. (5) 

4. D 41 (5) re = 2. (6) 

5. Substitute this value of re in (1) 10 — 3 y = 19. (7) 

- 3 y = 9, or y = - 3. (8) 


6. The common solution is re = 2, y = — 3. 

Check. Substitute in (1) 10 + 9 = 19. 

Substitute in (2) 14 — 12 = 2. 

Notice. — 1. That the coefficients of re and y were not the same in 
equations (1) and (2). That equations (1) and (2) were multiplied by 4 
and 3 respectively to make the coefficients of y the same except for 
sign, in equations (3) and (4). 

2. That the terms 12 y were made to disappear in Step 3 by adding 
equations (3) and (4). y is said to be eliminated by addition. 

3. If, in equations (3) and (4) the signs of 12 y were the same (for 
example, both +) then, in Step 3, we should have subtracted. 

Rule. — To solve a system of two simultaneous linear equa¬ 
tions having two variables by the addition or subtraction method 
of elimination: 

1. Multiply, if necessary, both the first and second equations 
by such numbers as will make the coefficients of one of the 
variables of equal absolute value. 

2. If the coefficients have the same sign, subtract one equation 
from the other; if they have opposite signs, add the equations. 

3. Solve the equation resulting from Step 2 for the other 
variable. 

4. Substitute the value of the variable found in Step 3 in any • 
equation containing both variables, and solve for the remaining 
variable. 

5. Check the common solution by substituting it in both of the 
original equations. 

* Read this “multiply both members of equation (1) by 4.” 


EQUATIONS HAVING TWO UNKNOWNS 207 


EXERCISE 113 

Solve the following systems of equations by the addition or 
subtraction method of elimination: 


>•( 

2 x + y = 7 

16 -l 

' 15 c = 8 - 4 d 

CO 

II 

1 

H 

CO 

L 20 c - 12 d = 15 

2 1 

’ 5 a + 2 6 = 13 

16 . { 

r 8 a + 5 b = 26 

,3a + 6=7 

[ 3 a = 31 - 4 6 

3 1 

m — 3n = — 12 

17 j 

f 18 p + 25 q = - 20 

[ 3 to + 2 n = 19 

[ 12 p + 15 q = — 14 

4 -1 

[ 3 r -f 4 s = 6 

18. J 

\ i 2 + -f w = 8 

i r — 2 s = — 8 

1 

[fa— w = — 3 

6 -1 

[ 2 c - 3 d = 18 

19 . j 

Cf A =8 

[ 3 c + 2 d = 14 

[f H + f B = - 1 

6. < 

f 7t + 5w = 3 

1 

20. j 

f 5 z + 6 y = 1.97 

l 2 t+ w = 0 

[ 4 z — 5 y = .4 

7. < 

f 10 z — 3 y = —7 

21 i 

5 +2* = - 5 

l 4 x + 5 y = 22 

1**+** = -H 

8. < 

f 9 r + 7 5 = 17 

22. | 

f 6a + 106 = - 1.9 

l 6 r — 5 5 = - 8 

[ 9a + 1006 = 15 

9. 

f 8 t + 18 w = - 15 

23. | 

[ 4 x + 3 y = 0 

1 St - 4m = - 11 

[ 5 a; + 4 y = — 1 

10. 

f 16 to + 3 n = 11 

24. J 

f 2 M = 5W 

l 20 to = 30 - 7 n 

[61 = 7W + 16 

11. 

f 11 x = 8y + 13 

25. 

f .5 A — B = - 8 

l 5x = 7 y + 16 

1 1.5 H — 2 B = — 12 

12. 

J 2 r = 3 s 

26. 

f c + .8d = 10 

\ 10 r + 6 s = 7 


[.5 c - d = -2 

13. 

j 4 1 + 12 w = 7 

27. < 

[ a r - .2 5 = 6 

1 6 t — 8w = — 9 

[ .3 r - .1 5 = 3.9 

14. 

J 5 m + 11 n = 47 

28. < 

f.3w--.7» = 27.4 

j 6 TO + 5 71 = — 1 

[ .2 w — .5y = 5 


208 


ALGEBRA 


149. Elimination by substitution. 

Example. Solve the system of equations 

I 7 x - 9 y = 15 ‘ (1) 

1 8 y - 5 a; = - 17 (2) 


Solution. 1. Solve equation (1) for x in terms of y: 

7x=15+9 V, x = 15 + 9 -g . 

2. Substitute this value of x in equation (2). 


Then 


8 V 


5(15 -j-9 y) _ _ 17 
7 


3. M 7 (4) 

4. 

5. C. T.; An 

6. D u 


(3) 


(4) 


56 y — 5(15 + 9 y) = - 119. 

. 56 y - 75 - 45 y = - 119. 

11 y = - 44. 
y = - 4. 

7. Substituting this value of y in equation (1), 

7 z + 36 = 15; 7 x = -21; x = -3. 

8. the common solution is x = — 3, y = — 4. 

CM;. In (1), does - 21 + 36 = 15? Yes. 

In (2), does - 32 + 15 = - 17? Yes. 

Note. — In Step 3, when multiplying — —^ ^ by 7, we obtain 
r■ 5(15 ± - 9 . ^ , or 5(15 + 9 y). 


Rule. — 1. Solve one equation for one variable in terms of 
the other variable. 

2. Substitute for this first variable in the other equation the 
value found for it in Step 1. 

3. Solve the equation resulting in Step 2 for the second 
variable. 

4. Substitute the value of the second variable, obtained in 
Step 3, in any equation containing both variables and solve for 
the first variable. 

6. Check the solution by substituting it in the original equa¬ 
tions. 






EQUATIONS HAVING TWO UNKNOWNS 209 


•■1 

' 3 x + 2 y = 16 

16 ' 1 

' 5t + 25 = 1 

. 7 x + y = 19 

; 121 +- 5 s = 1 

2 -1 

a — 5 b = — 2 

16 - i 

’ 16 M + 9 N = 24 

i, 9 a + 7 b = 34 

k 4iU +-3 N = 7 

3 J 

f 3c + d = 6 

n. { 

' 12 X- Y = 14 

1 

[ 7 c 5 d = — 2 

. 18 X + 5 Y = - 5 

4. | 

[3r + 7s = - 5 

18. j 

r 5c + lOd = 7 

1 

i 6 r - 5 s = 28 

,7c - 2d = 5 

6. < 

f 9t + 4w = - 7 

19. { 

4 p — 5 q = 4 

1 11* +8w = 7 

[8p — 15 g = 2 

6. < 

\2x + llz = l 
[z + 7 -x =0 

20. | 

[ 10 n - 11 2 = - 4 
[ 5 n + 5 z = 4 

7. < 

f 10 A = 3 B — 1 

2 i. | 

[ 16 £ +- 9 w =8 

l 7i + 25 =28 

[ 8 x — 3 w = 9 

8. 

f 9 m — 14 n = — 8 
l m = 2 n 

22. ] 

f 5 z — 3 y = 3 
l 40 z + 6 y = 3 

9. 

f 12 p - 5 g = - 6 

23. < 

f 50 = 25 

l 8 p +- g = — 30 

110 Z> - 6 = 150 

10. < 

f 18 w — 4 r = —7 

24. < 

\2W + 90 = -3 

[ r — 6 w = 1 

[5JU + 60 = - 13 

11. < 

f 13 x - 12 t = 35 

1 x = 9 t 

25. < 

f 8 v +- t = 8 
l 6 v +- ±t = 3 

12 - 1 

f9x+-4i/= — 21 

26. < 

f 3 r = 4 & — 9 

[ 7 x — 3 y = 2 

l f r +- 2 & =0 

13 ‘ 1 

[ 7 r + 8 5 = 36 

27. | 

r .2 X 4- .5 Y = 12 

[ 11 r + - 4 

i 3 X - 7 = 10 

14 ' 1 

f 3 a +- 2 6 = 1 

28. | 

' 10 X + = 5.5 

t 12 a - 10 b = 13 

k 5 5 + 3# = 11 


Historical Note. — The earliest use of the Substitution Method of 
Elimination in print, of which we have any record, is in Newton’s 
Arithmetica Universalis, in 1707. 


210 


ALGEBRA 


150. Simultaneous fractional equations. 


Example. 


Solve the system of equations 

JL _3_ = o 

* +3 y + 4 
x(y -2) - y(x - 5) = - 13 


( 1 ) 

( 2 ) 


Solution. 1. Simplify equation (1) 


x + 3 y + 4 

M(a;+ 3 )(l/+ 4 ) 7(j/ + 4) — 3{x + 3) = 0. 

7*/ + 28-3z-9 = 0. 

7 y — 3 s + 19 = 0, or | 3 a; - 7 y =~19~| (3) 

2. Simplify equation (2) 

x(y - 2) - y(® - 5) = - 13. 
xy — 2 x — xy + 5 y = — 13. 


- 

2 x + 5 y = — i3, or | 2 x — 5 y = 13 | 

(4) 

3. Therefore equations (1) and (2) become equations (3) and (4). 



3 x — 7 y = 19. 

(3) 


2 x — 5 y = 13. 

(4) 

4. M 6 (3) 

15 x — 35 y= 95. 

(5) 

M 7 (4) 

Ux-35y= 91. 

(6) 

5. (5)-(6) 

4. 



6. Substitute in (4) 8 — 5 y = 13; — 5 y = 5; y = — 1. 
Check by substituting in (1) and (2). 


Equations (3) and (4) of this solution are called the standard 
form of equations (1) and (2). 

Rule. — To solve complicated simultaneous linear equations 
having two variables: 

1. Reduce the equations to standard form by clearing of 
fractions (when necessary) and simplifying. 

2. Solve the resulting system of equations by either method 
of elimination. 

3. Check by substitution in the original equations. 











EQUATIONS HAVING TWO UNKNOWNS 211 


EXERCISE 115 


Solve the following systems, 
traction method of elimination, 
method: 

3 x _ 4j/ = _ 

2 3 

2 x _ y _ 7_ 

~3 4 “ 12 

a -(- 3 6 

E 2 


1 . 


3d - 


4 c + 


6+3_ 

~4 

4 


3 

2 

3 

’ 2 


6 

2d - 3 


= 13 


= -7 


m — 1 

w + 6 

m — 5 _ 

_ 2 

n + 1 

3 

> 5 _ 

5 

6 4 ” 

12 

3 + 2 r 

7 +3^ 

5 

10 

5 x — ^ 
2 

II 

1 

to | l— 1 

4 £ + 1 

_ y - 4 

6 

4 

32-1 

2 w + 1 

4 

2 

32 + 1 

2 w - 1 

2 

3 


= 1 


= 1 


some by the addition or sub- 
and some by the substitution 


2 t + « 


10 . 


11 . 


12 . 


13 . 


14 . 


1 

t — s 2 
4 - f - s = 2 
21 5 

A + B - 2 _ 1 
A - B 5 
A + 2 B - 3 _ 


1_ 

'8 


A -3B 

3 x —4 y _ x + 8 y _^ 


2 5 

2 x + y _ 4x -3 y = Q 


5 9 

w — 2 _ 2 + 2 


w + 1 2 + 1 

w — 3 _ z + 5 
w — 4 2 + 4 

a - 36 - 1 


2a - 36 - 1 
5 - a + 66 


= 1 


= 1 


7a - 96 - 1 
X — Y Z + F. 


4 

2 X — Y 

6 

3 £ — 1 


5 

2i+y 

3 

2r - 1 


= 1 


= 0 


2 1 
3 t + 1 2Q + 1) _ Q 




















































212 


ALGEBRA 


151. When solving problems by means of two or more literal 
numbers, as many independent equations must be obtained 
from the conditions of the problem as there are letters used. 

Note. — Most problems that can be solved by means of equations 
having two unknowns, can also be solved by an equation having only 
one unknown. Nevertheless, use two unknowns when solving the 
following problems. 

EXERCISE 116 

1. Find two numbers whose sum is 20 and whose difference 
is 7. 

Solution. 1. Let l = the larger number and s = the smaller 
number. 

(Complete the solution.) 

2 . Separate 39 into two parts such that 4 times the smaller 
exceeds the larger by 6. 

Solution. 1. Let l = the larger part and s = the smaller part 
of 39. 

2. I + s = ? 

3. Also 4 s -1= ? 

(Complete the solution.) 

3 . Solve Example 2 by an equation having only one unknown. 

4 . If 2 be added to both numerator and denominator of a 
certain fraction, the resulting fraction equals ; if 7 be added 
to both terms of the original fraction, the result equals §. What 
is the fraction ? 

Solution. 1. Let n = the numerator and 

d = the denominator of the fraction. 

—j = the fraction. 
d 

2Th »- tU-w (d 

3Ak0 ’ ItH- < 2) 

Reduce equations (1) and (2) to the standard form. (See page 236.) 
Complete and check the solution. 




EQUATIONS HAVING TWO UNKNOWNS 213 


5. Find two numbers whose sum is 56, and such that of 
the greater exceeds -J of the smaller by 3. 

6. Solve Problem 5, using only one unknown. 

7 . The larger of certain two numbers exceeds twice the 
smaller by 8. Five times the smaller exceeds twice the larger 
by 5. What are the numbers ? 

8. Separate 34 into two parts such that the quotient of the 
larger divided by the smaller is 3, and the remainder is 6. 

9 . If. 2 be added to the numerator of a certain fraction, 
the resulting fraction equals ^. If 1 be added to the denomina¬ 
tor of the original fraction, the resulting fraction equals -J. 
What is the fraction ? 

10. A’s age now exceeds twice B’s age by 3. Three years 
ago, x4 was 4 times as old as B was then. What are their ages 
now? 

11 . The perimeter of a rectangle is 72 inches. The width 
exceeds of the length by 1 inch. What are the length and 
width ? 

12. A mixture, made up of copper and tin, weighs 25 pounds. 
It consists of two parts copper to five parts tin. How many 
pounds of pure copper must be added to make the resulting 
mixture 50% copper ? 

13. Twice the shorter side of a rectangle exceeds the longer 
side by 6 inches. One half the longer side exceeds one seventh 
the shorter side by 9 inches. Find the sides. 

14. If three times the smaller of two numbers be divided 
by the larger, the quotient is 2 and the remainder is 4. The 
larger exceeds the smaller by one. What are the numbers ? 

15. Six years ago, A was 12 times as old as B was; three 
years from now, he will be 3 times as old as B is then. What 
are their ages now ? 


214 


ALGEBRA 


16 . A purse contained $5.45 in quarters and dimes. Pay¬ 
ment for a 5^ purchase was made with a quarter, and dimes 
were received in change. Then the number of dimes was twice 
the number of quarters remaining. How many dimes and 
quarters were there at first ? 

17 . If seven times the smaller of two numbers be divided by 
the larger, the quotient is 1 and the remainder is 3. The larger 
exceeds 5 times the smaller by one. Find the numbers. 

18 . Separate 87 into two parts such that the quotient of the 
larger divided by the smaller is 5 and the remainder is 3. 

19 . If the numerator of a fraction be doubled and the de¬ 
nominator be increased by 5, the resulting fraction equals -J. 
If the numerator be decreased by 1, and the denominator be 
increased by 7, the resulting fraction has the value What is 
the given fraction ? 

20. The length of a room is 7 feet more than its width. 
If the room is made 1 foot wider and 3 feet longer, the area is 
increased by 66 square feet. Find the original dimensions. 

Solution 1. Let Z = the no. of feet in the length; 

2. and w = the no. of feet in the width. 

3. 

4. 


Z = w + 7 


( 1 ) 



Length 

Width 

Area 

Old dimensions 
New dimensions 

Z 

Z —3 

W 

W + 1 

lw 

(Z + S)(w + 1) 


5. 


.'. (Z 3)(w -j- 1) — Iw -h 66 
(Complete the solution, using equations (1) and (2).) 


( 2 ) 


21. If 5 feet be added to the width and 15 feet to the length 
of a lot, the area is increased 1750 sq. ft. If 10 feet be subtracted 
from the length and 15 feet be added to the width, the area is 
increased 750 sq. ft. Find the length and width. 












EQUATIONS HAVING TWO UNKNOWNS 215 


22. One sum of money is invested at 6%. A second sum, 
which exceeds the first sum by $150, is invested at 7%. The 
interest on the second sum exceeds the interest on the first sum 
by $19. What are the two sums ? 

23 . How many pounds each of 35^ coffee and 55^ coffee must 
be mixed to make a mixture of 100 pounds to sell at 40^ per 
pound ? 

24. The total value of two investments is $4500. The annual 
interest earned on one is 5%, and on the other is 6%. The 
total interest is $252. What are the two sums? 

25. The simple interest on $1200 at 7% for a certain number 
of years exceeds the simple interest on $750 at 6% for a second 
period of years by $27. The second period of years exceeds 
the first by 2. How long is each period of years ? 

26 . How many pounds each of nuts worth 18j£ per pound 
and 30^ per pound must be mixed to make a mixture of 75 
pounds to sell at 22^ per pound ? 

27. It took a motor boat 1-J hr. to go 20 miles downstream, 
and 2f hr. to return. What was its own rate in still water and 
what was the rate of the current of the river ? 

28. Some boys were rowing on a river. It took them only 
three hours to go 20 miles downstream; but it took 10 hours 
to return. What was their own rate of rowing in still water 
and what was the rate of the current ? 

29. On a river whose current is known to be miles per 
hour, some boys rowed downstream for 2 hours. It took them 
6 hours to return. How far did they go, and what was their 
rate of rowing in still water ? 

30 . If a certain lot is made 5 feet longer and 2 feet wider, its 
area is increased 415 square feet. If it is made 3 feet narrower 
and 10 feet longer, its area is decreased 60 square feet. What 
are its dimensions ? 


216 


ALGEBRA 


152. Relations among digits of a number. Integers are 
written by means of the digits 0, 1, 2, 3, . . . 9. 

Thus, 38 is a number of 2 digits. 3 represents 3 X 10 units; 8 repre¬ 
sents 8 units; the total is 38 units. The sum of the digits is 3 + 8 or 11. 

If t is the tens’ digit, and u is the units’ digit, the number contains 
10 t + u units. The sum of the digits is t + u. 

When the digits are reversed, a new number is formed. Thus, reversing 
the digits of 52 gives 25. 

Notice that 52 = (5 X 10 + 2) units, 
and that 25 = (2 X 10 + 5) units. 

Similarly, if x and y are the tens’ and units’ digits of a number, then 
the number contains 10 x + y units. If the digits are reversed, the 
new tens’ digit is y and the units’ digit is x ; then the new number con¬ 
tains (10 y + x) units. 


EXERCISE 117 

1. How many units are there in the number: 

a. whose tens’ digit is 8 and units’ digit is 5 ? 

b. whose tens’ digit is a and units’ digit is b ? 

c. whose tens’ digit is b and units’ digit is a? 

2. The hundreds’, tens’, and units’ digits of a certain number 
are x, y, and z respectively. 

a. How many units does this number contain ? 

b. If the digits are reversed, how many units does the new 
number contain ? 

3. Let t represent the tens’ digit and u the units’ digit of a 
certain number of two digits. 

a. Represent the difference of the digits. (See § 94, page 
118 .) 

b. Represent the number itself. 

c. Represent the quotient of the tens’ divided by the units’ 
digit. 

d. Represent the number with digits in the reverse order. 

e. Represent this last number divided by the sum of the 
digits of the given number. 


EQUATIONS HAVING TWO UNKNOWNS 217 


4. The sum of the digits of a certain two-digit number is 
11. If the digits be reversed, and the new number be divided 
by the old, the quotient is 2 and the remainder is 7. What is 
the number? 

Solution . 1 . Let t — the tens’ digit of the number, 

and u = the units’ digit of the number. 

.*• t + u = 11. (1) 

2. Then (10 u + t) = 2(10 t + u) + 7. (2) 

(Complete and check the solution, using equations (1) and (2).) 

5. The tens’ digit of a certain two-digit number exceeds 
its units’ digit by 5. If the digits be reversed, the new number 
is one fourth of the result obtained by adding 3 to the given 
number. What is the number ? 

6. The tens’ digit of a certain two-digit number exceeds 
twice the units’ digit by 1. If the digits be reversed, the sum 
of the new number and the original number is 143. What is 
the number? 

7. The sum of the two digits of a certain two-digit number 
is 9. If the digits be reversed, the resulting number is 9 less 
than 3 times the original number. What is the number? 

8. A certain number consists of two digits. If the digits 
be reversed, the sum of twice the original number and the new 
number is 168. Also the original number exceeds 4 times the 
sum of its digits by 3. What is the number? 

9. A certain number consists of three digits of which the 
units’ digit is 5. The sum of the digits is 12. If the hundreds’ 
and tens’ digits are reversed, the new number is 90 less than 
the original number. What is the number? 

10. In a certain number consisting of three digits, the tens’ 
digit is 6. The hundreds’ digit is 1 less than twice the units’ 
digit. If the hundreds’ and units’ digits be interchanged, the 
new number exceeds one half the original number by 85. What 
is the number? 


218 


ALGEBRA 


153. Fractional equations, linear in i and 

10 - § = 8 . 


Example. Solve the system 


Solution. 1. Ms(l) 
2. M 3 (2) 


x y 

8+“=-l. 

x y 


— — = — 3. 


^ = 37. 

X 

74 = 37 x, or x = 2. 


3. Adding, 

4. 

5. Substitute 2 for x in (1) 5 — - = 8. 

y 

6. Solving for y, y = — 3. 


EXERCISE 118 


Solve the following systems of equations : 


- + -= 5 
x y 


= 3 


12 _ 6 = _ i 

x y 

- + —=-8 


5 _ 8 = i 
x y 

§ + ?=_ 13 
x y 


15 . 
B 


7 
’ 2 


7 , 5 = _3 
A B 4 


i + 3 - = 7 
x y 

10 _ 5 = _ 15 
x y 2 

2y- 1 - = l 

3 y+~=-1 

X 


1 = 9 


+ -=- 1 
s 


- +1 = - 10 
a b 


25 

2 


( 1 ) 

( 2 ) 

(3) 

(4) 

(5) 











EQUATIONS HAVING TWO UNKNOWNS 219 



Work problems and problems about reciprocals 


Note. — For work problems, review Examples 1-4, page 176. 

The reciprocal of 3 is 1 3 or £; of f is 1 -s- f or §; of x is 

11. The sum of the reciprocals of certain two numbers is -J. 
Twice the reciprocal of the smaller, diminished by the reciprocal 
of the larger, also is -J. What are the numbers? 


Suggestion. 


If l = the larger, then | is its reciprocal. 


12. There are two unequal numbers such that the reciprocal 
of the smaller, diminished by the reciprocal of the larger, is •§. 
Twice the reciprocal of the smaller plus five times the recip¬ 
rocal of the larger equals 6. What are the numbers ? 

13. A and B together do a certain piece of work in 10 days. 
They can do it also if A works 12 days and B works 6 days. 
How long would it take each to do it alone? 

Suggestion. — If A can do it in a days, how much can he do in 1 day ? 

14 A and B together can do f of a certain piece of work in 
3 days. They can do all of it if A works 9 days alone and B 
works 2 days alone. How long would it take each of them to 
do all of it alone ? 

15. A workman and one helper did a certain piece of work 
in 6 days. On another occasion, the workman and two helpers 
did the same work in 4 days. If the two helpers worked at 
the same rate, how long would it take the workman alone or 
each of the helpers alone to do the work ? 

16. The sum of the reciprocals of two numbers is If 
twice the reciprocal of the smaller be diminished by eight times 
the reciprocal, of the larger, the result is — 7. Find the numbers. 




220 


ALGEBRA 


154. Literal simultaneous equations. 


Example. Solve for x and y the system: 



f ax -f by = c 

(1) 


\ rx + sy = t 

(2) 

Solution. a. 

Eliminate y. 


1. M, (1) 

asx + bsy = cs. 

(3) 

2. M 6 (2) 

brx + bsy = bt. 

(4) 

3. (3) - (4) 

asx — brx = cs — bt. 

(5) 

4. Factoring, 

x(as — hr) — cs — bt. 


5. D (as-6r) 

cs — bt 

x ~ u ' 

as — br 


6. Next, going back to (1) and (2), eliminate x. 


6. M r (1) 

arx + bry = cr. 

(6) 

7. Ma (2) 

arx + asy = at. 

(7) 

8. (6) - (7) 

bry — asy = cr — at. 

(8) 

9. Factoring, 

y(br — as) = cr — at. 


10. D(ft r _a S ) 

cr — at __ at — cr 

y — »or y — ■ 

br — as as — br 



11. the common solution is x = ££- ^1) y = — -— • 

as — or as — or 


The results in Step 11 are formulae for the two numbers x 
and y, which satisfy the given equations. For every set of 
values of a, b, c, r, s, and t, there is a definite value of x and of 
y, which can be obtained by substituting in Step 11 the values 
selected for a, b, c, r, s, and t. 

EXERCISE 119 


Solve the following systems for x and y: 


^ \2 x — 3 y = a 

4 | bx + y = b 

[3x +4y = b 

[ ax + 2 y = a 

2 f cx + cy = 2 a 

6 \ ax -{-by = 1 

\ cx — cy = 2 b 

1 x + y = 1 

3 (4 x + ay = b 

6 f ax — by = 1 

\ 5 x + by = a 

\ mx + ny = 1 







EQUATIONS HAVING TWO UNKNOWNS 221 


7. 

8 . 

9. 

10 . 


i 

“■ { 


x = y + a 
2 x - 3 y = b 

x = y — a 
ax + y = b 

cx = y 
x + ay = b 

y = mx 
sx + ry = t 

ax — by = c 
dx - ey = f 

2x + y = 3 a + b 
x +2y =3 a + 2 6 


13. 


14. 


15. 


16. 


17. 


18. 


2x +3y = 5a + b 
3 x + 2 y = 5 a — b 

mx y = m 2 + n 
x + my = m + mn 

f rx + ry = 2 r 2 
l so; — s?/ = 2 s 2 

foe + ay = a -f b 
abx + = cl 2 + b 2 

ax = by 

abx + aby = {a + b) 2 

bx — ay = ab 
3 ax + 3 by = 2 a 2 — b 2 


19. a. Find two numbers whose sum is m and difference is n. 

b. Using the results as formulae, find the numbers whose sum 

is 33 and difference is 7. 

20. a. Separate n into two parts such that the larger will 
exceed twice the smaller by m . 

b. Using these results as formulae, find the two parts when 
n is 50 and m is 8; also when m = 11 and n = 97. 

21. a. Separate a into two parts such that when the larger 
is divided by the smaller the quotient is b and the remainder 
is c. 

b. What are these two parts when a = 75, b = 2, and c = 3 ? 

22. a. A is now r times as old as B. s years ago, A was t 
times as old as B. What are their present ages ? 

b. What are these ages when r = 2, s = 15, t = 5? 

23. The length of a certain rectangle exceeds its width by 
c feet. The perimeter is p feet. Find its length and width. 

Note. — Exercise 167, page 290, can be done now. 


222 


ALGEBRA 


OPTIONAL TOPIC 

155. Dependent equations are two equations which have 
exactly the same solutions. For all practical purposes they are 
the same equation. One can be obtained from the other by 
multiplying it or by dividing it by the same number. 

Thus, x + V = 10 and 2 x + 2 y = 20 ' are dependent equations. 
x = 3, y = 7 is a solution of each. Any other solution of the one is 
also a solution of the other. 


156. If a system consists of two dependent equations, the sys¬ 
tem cannot be solved, because actually there is only one equation. 

Thus, consider L 

\2z + 2?/ = 20 (2) 

If we try to eliminate by substituting (10 — y ) for x in (2), we get 
2 x + 20 — 2 x = 20, or 0 = 0. x cannot be determined. 

157. Inconsistent equations are two independent equations 
which do not have any common solution. 

Thus, x + 2 y = 7 and 4 x + 8 y = 40 are inconsistent. 

If 4 x + 8 y = 40 is divided by 4, we get x + 2 y = 10 
Clearly, if £ + 2 y = 7, x + 2 y cannot = 10. 

If we try to solve the system / x + 2 ?/ — 7 

[ 4 x + 8 y = 40 

by the subtraction method, we get I f X y ~ 

J 5 & 1 4 z + 8 ?/ = 40. 

Subtracting, 0 = — 12. This is impossible. 


EXERCISE 120 

1. a. Draw the graphs of 2 x — y = 4 and 6 x — 3 y = 12. 
b. What happens ? How many common points do the graphs 

have? How many common solutions , then, do the equations 
have ? c. What kind of equations are these ? 

2. Similarly, study equations x + 2 y = 6 and 2 x + 4 y = 12. 

3. a. Draw the graphs of 3 x — y = 5 and 6 x — 2 y = 18. 
6. What kind of lines are these graphs ? Do these equations 

have a common solution ? c. Are these equations independent ? 
d. Are these equations simultaneous ? e. What kind are they ? 

4. Similarly, study equations 2 x — 3 y = 6 and 4 x — 6 y = 30. 


EQUATIONS HAVING TWO UNKNOWNS 223 


REVIEW EXERCISE X 


1. Divide 4m 4 -9m 2 + 6m-l by 2m 2 + 3m-l. 

2. Reduce to lowest terms 3 C ~ 6 c ; also — ~ ~ 10 

6c 2 -24’ 2 z 2 — 50 

3. Solve for k : ^ - i- (ifc _ 11) = ? (it - 25) + 34. 

5 14 7 

4. a. Solve T = 2ttR(R + #) for H. 

b. In your result, let tt =3.14, R = 10, T = 794.42. Find 
H correct to tenths. 


5. a. The three digits of a number are a, b, and c. Repre¬ 
sent the number formed by reversing the digits. 

b. A man earns e dollars a month, and spends s dollars a 
month. How much will he have saved in 3 years ? 

c. John is y years old now. Write the equation showing that 
4 times his age 3 years ago is twice his age 5 years from now. 


6.* a. Simplify — - 1 - - 
or — a — o 


a — 4 


4 a -j- 3 


CL -f- 3 
a 2 + a — 2 


b. Check the solution by letting a = — 1. 


7. Simplify 


(2 r — 3 s) 2 4 r 2 — 4 


3 rs — 3 


(r + 2 s) 2 4 r 2 — 9 s 2 5 r 2 + 10rs 


8. Solve and check — - - x - ^7°— -j- — 

2 a: 2 — 2 x x 2 — 1 


3 # 

z 2 - r 


9. Draw the graph of y = mx + b when m = 3 and 6 = 5. 

10. Draw the graph of y = \ x 2 . 

11. The following table gives the number of clear days in 
Chicago in each month of a certain year: 


Month 

Jan. 

Feb. 

Mar. 

Apr. 

May 

Ju. 

JUL. 

Aug. 

Sep. 

Oct. 

Nov. 

Dec. 

Clear days 

12 

7 

8 

10 

10 

13 

12 

19 

18 

15 

4 

3 


Represent these numbers by a broken-line graph. 



































XIII. SQUARE ROOT AND RADICALS 


158. The square root of a perfect square monomial was 
found by inspection in § 89, page 108. Review this section. 

159. Every number has two square roots. They have equal 
absolute values but opposite signs. 

Thus, V9 a A b 2 = + 3 a 2 b, because (+ 3 a 2 b) 2 = 9 a A b 2 . 

Also, V9 a A b 2 = — 3 a 2 b, because (— 3 a 2 b) 2 = 9 a 4 6 2 . 

These two roots are written together by means of the double 
sign +, read plus or minus. 

Thus, V9 a% 2 = ± 3 a 2 b; read plus or minus 3 a 2 b. 

Of these two square roots, the positive one is called the prin¬ 
cipal square root. When the square root is mentioned, usu¬ 
ally the principal square root is meant. 

160. The square root of a large number often may be obtained 
by inspection, after factoring the number. 

Rule. — The square root of the product of two or more positive 
numbers equals the product of their square roots. 

Thus, V4 • 25 = 2 • 5 = 10; Vab =Va- Vb. 

V1764 a 4 = V4 • 441 a 4 = V4 • 9 • 49 a 4 = 2 • 3 • 7 • a 2 = 42 a 2 . 


EXERCISE 121 


Find the 

36 to 4 

square roots of: 
7. 49 ad 

13. 

196 A 4 

19. 

576 a 2 

64 a 6 5 2 

8. 400 a 2 

14. 

225 x 2 

20. 

2500 b 4 

121 m 2 n 

4 9. 121 ad 

15. 

900 a 2 

21. 

1089 x 4 

169 x 6 

10. 256 y 4 

16. 

169 a 2 6 4 

22. 

1764 y 6 

81 c 8 d 2 

11. 169 z 2 

17. 

196 x s 

23. 

1936 zV 

144 aY 

12. 625 r 4 

18. 

224 

225 y % 

24. 

2304m 6 » 4 








SQUARE ROOT AND RADICALS 


225 


25. Find the square roots of 9 a 2 — 12 ax + 4 x 2 . 

Solution. 1. V9 a 2 - 12 ax + 4 x 2 = V (3 a - 2 x) (3 a - 2 x). 

2. /. V9 a 2 - 12 ax + 4 x 2 = V(3 a - 2 x) 2 

= =*= (3 a - 2 x). 


26. 

25 a 2 

- 10 a + 1 

36. 

X 2 

+ 

f x + ih 

27. 

c 2 - ] 

16 c + 64 

37. 

y 2 

- 

iv + i 

28. 

m 2 - 

20 mn + 100 n 2 

38. 

z 2 

- 

T z +T5T 

29. 

4 a 2 - 

- 20 a + 25 

39. 

k 2 

+ 

k + i 

30 

9 a 2 - 

- 30 ab + 25 b 2 

40. 

X 2 

+ 

i x + i 

31. 

16 c 2 

- 24 cd + 9 d 2 

41. 

w 2 



32. 

36 x 2 

-f 60 xy + 25 y 2 

42. 

TO 2 


!»+-r 

33. 

49 m) 

1 — 70 mn + 25 n 2 

43. 

X 4 

- 

1 x*y + A y 2 

34. 

144 a? + 24 a + 1 

44. 

y 2 

+ 

A yt + tIt * 2 

35. 

36 x 2 

— 84 xy + 49 y 2 

45. 

X 2 

- 

I “ + M s 2 


161. Making perfect square trinomials. 

Example. Make a perfect square trinomial of 
25 x 2 — 30 xy + (?) by supplying the missing term; then find 
the principal square root of the result. 

Solution. 1. In the expressions of Examples 25-45 above, ob¬ 
serve : 

Two terms (the first and third) are perfect squares, preceded by plus 
signs; the remaining term is twice the product of their square roots. 

2. In this example, V25 x 2 = 5 x. 

3. —30 xy must be 2 • (5 x) • some number. 

.*. that number must be 3 y, since 30 xy -s- 10 x = 3 y. 

4. The remaining term must be (3 y) 2 or 9 y 2 . 

5. The perfect square trinomial is 25 p 2 — 30 xy + 9 y 2 

6 . and V25 x 2 — 30 xy + 9 y 2 = (5 x — 3 y). 

[Examples for this section appear on page 226.] 







226 


ALGEBRA 


EXERCISE 122 


Complete the square and give the principal square root. 


1. 

a 2 + 6 a + (?) 

16. 

a 2 — 11 a +(?) 

2. 

r 2 - 12 r+(?) 

17. 

r 2 + *+(?) 

3. 

t 2 - 18*+(?) 

18. 

y 2 - *y+(?) 

4. 

m 4 — 24 m 2 + ( ?) 

19. 

2 2 -**+(?) 

5. 

36 x 2 — 12*+(?) 

20. 

m 2 — f m + (?) 

6. 

a: 2 -|- 22 x + (?) 

21. 

* 2 +t*+(?) 

7. 

a 2 - 16 a +(?) 

22. 

w 2 — f w + (?) 

8. 

x 2 — 10 xy +(?) 

23. 

a 2 — fa + (?) 

9. 

y 2 ~ 8yz +(?) 

24. 

x 2 + ix+W 

10. 

a 4 - 20 a 2 b + (?) 

26. 

y 2 -iy+0) 

11. 

c 2 - 14 c(P+(?) 

26. 

z 2 ~iz+(?) 

12. 

x 2 — 3 x + (?) 

27. 

m ,2 + |m+(?) 

13. 

y 2 + 5y+(?) 

28. 

£ 2 — T £ + (?) 

14. 

Z 2 +7z+(?) 

29. 

2 / 2 - iy +(?) 

16. 

m 2 — 9 m + (?) 

30. 

z 2 - fz +(?) 


EXERCISE 

123 


Give the principal square root of: 


1. 

49 m 6 

6. 

16 a 2 - 24 a + 9 

2 . 

144 a 4 6 2 

7. 

4 z 2 — 20 xy + 25 y 2 

3. 

225 r 6 s 2 

8. 

169 a 2 - 26 a + 1 

4. 

16 a 10 

25 5 4 

9. 

m 4 — 10 + 25 n 2 

6. 

4 

10. 

a 4 - 6 a 2 b 2 + 9 ¥ 

169 # 4 


Complete the square and give the principal square root: 


11. 

s/ 4 - 12 /+(?) 

16. 

(x + 3) 2 

19. 

t 2 - 

**+(*) 

12. 

z 4 - 3z*+(?) 

16. 

(2 x - 5) 2 

20. 

(y - 

. 3^2 

5/ 

13. 

wj 2 — 5 w -}-(?) 

17. 

(* - |) 2 

21. 

y 2 ~ 

f2/+(?) 

14. 

(3 m- l) 2 

18. 

(x + # ) 2 

22. 

(2 z 

-i) 2 




SQUARE ROOT AND RADICALS 227 


162. f The square root of a perfect square polynomial can be 
found by a long division process. 

Example. Find the square root of 9 x 2 + 25 y 2 — 30 xy. 


Solution. 1. Arrange the polynomial in descending powers of x. 


9x 2 
9 x 2 


2. V9 x 1 = 3 x. Write 3 x in the root. Root 3 x 

3. Square 3 x, getting 9 x 2 . 

Write it below 9 x 2 . 

4. Subtract, obtaining the first 

remainder. 

5. Trial divisor, 2 • 3 x = 6 x. 

— 30 xy -i- 6 x = — 5 y. 

Write — 5 y in the root and 
add — 5 y to 6 x, forming 
the complete divisor. 

6. Multiply the complete divisor 

by - 5 y. 

7. Subtract. In this case, no remainder. 

8. The square roots are + (3 x — 5 y) and — (3 x — 5 y). 


- 5y 


30 xy + 25 y 2 


6 x 


- 5 y 
6 x — 5 y 


— 30 xy + 25 y 2 


— 30 xy + 25 y 2 


EXERCISE 124 


Find the square root of: 

1. 25 z 2 + 40 xy + 16 y 2 

2 . 9 a 2 - 30 ab + 25 b 2 

3. 36 c 2 — 60 cd + 25 cP 


6. 4z 4 + 4z 3 + 5x 2 + 2z + l 

7. 4 x 4 - 4 z 3 + 5 z 2 — 2 * + 1 

8 . 9 z 4 + 6 z 3 + 13 z 2 + 4 x + 4 


4. 49 m 4 — 28 m 2 n + 4 n 2 9. 9 a 4 + 12 a 3 — 2 a 2 — 4 a + 1 
6. 25 r 2 - 70 rs + 49 s 2 10. y* - 6 y* + 13 y 2 - 12 y + 4 

11 . 4 c 4 - 12 cH + 17 c 2 ^ 2 - 12 cd? + 4 d 4 

12. 9 m 4 — 24 m 3 n + 10 m 2 n 2 + 8 ran 3 + n 4 

13. 1 -2x - x 2 + 2x z +x 4 


14. a 2 + 6 2 + c 2 + 2 ab + 2 ac + 2 be 

15. 4 a 2 + 9 b 2 + c 2 + 12 ab — 4 ac — 6 be 

16. 9 a 2 — 24 ab + 30 ac + 16 b 2 — 40 be + 25 c 2 

17. 4 z 2 - 12 xy - 16 xz + 9 y 2 + 24 yz + 16 z 2 
^ Note. — Not required by Regents or C.E.E.B. 









228 


ALGEBRA 


163. The square root of an arithmetical number. 

Any number having one or two figures to the left of the 
decimal point (like 1, or 99.5) equals or exceeds 1 and is less 
than 100; therefore its square root, equal to or exceeding 1 and 
less than 10, must have one figure to the left of the decimal point. 

Any number having three or four figures to the left of the 
decimal point (like 100.25 or 9999.7) equals or exceeds 100 and 
is less than 10,000. Therefore its squqre root, equal to or ex¬ 
ceeding 10 and less than 100, must have two figures to the left 
of the decimal point. 

Hence, if the given number is divided into groups of two 
figures each, starting at the decimal point, for each group in 
the number there will be one figure in the square root. The 
groups are called periods. 

Thus, 2345 becomes 23 45. It has two periods. Its square root 
has two figures to the left of the decimal point, a tens’ figure and a 
units’ figure. 34,038 becomes 3 40 38. It has three periods. Its 
square root has three figures to the left of the decimal point. The left¬ 
most period may contain only one figure, as in this case. 

A decimal is divided in the same manner, counting in both directions 
from the decimal point. Thus, 34256.895 becomes 3 42 56.89 50. 
The square root of this number has three figures to the left of the 
decimal point and two to the right. 

The square root is found as follows: 

Example 1. Find the principal square root of 4624. 

Solution. 4624 must be the square of a number more than 60 
and less than 70, since 60 2 = 3600 and 70 2 = 4900. 

That is, the square root is 60 + some number less than 10. 





a 


b 




60 + 8. 


6 8. 

1 . 60 2 = 3600. Subtract from 4624. 


46 24. 


46 24. 

2. 

Trial divisor, 2 X 60 = 120. 


36 00 


36 

3. 

1024 -h 120 = 8+ Add 8 to 120. 

120 

10 24 

120 

10 24 


Add 8 to 60 in the root. 

8 


8 


4. 

Multiply 128 by 8. Subtract. 

128 

10 24 

128 

10 24 


Note. — Use only the form b when writing the solution. 









SQUARE ROOT AND RADICALS 


229 


Rule. — To find the principal square root of a number: 

1. Separate the number into groups (periods) of two figures 
each, starting at the decimal point, and forming the groups each 
way from the decimal point. 

2. Find the largest square number not more than the left-most 
period. Write its square root as the first figure of the square 
root. Subtract the square number itself from the first period. 

3. To the remainder, annex (bring down) the next period. 

4. Form the trial divisor by doubling the root already found, 
and annexing a zero. 

6. Divide the remainder formed in Step 3 by the trial divisor. 
Annex the quotient to the root already found, and add it to the 
trial divisor to form the complete divisor. 

6. Multiply the complete divisor by the new figure of the root. 
Subtract the result from the remainder formed in Step 3. 

7. If other periods remain, repeat Steps 4, 5, and 6 until there 
is not a remainder, or until the desired number of decimal 
places for the root have been obtained. 

Note. — In Step 6, sometimes the product is greater than the re¬ 
mainder. In such cases, the last figure obtained for the root is too 
large. Substitute for it the next smaller integer. 

Example 2. Find the principal square root of 5207.0656. 


Solution. 1. The largest square number in 52 
is 49. Write V49 or 7 in the root. Subtract. An¬ 
nex 07. 


7 2.1 6 
52 07.06 56 
49 

2. Trial divisor is 2 X 7. Annex 0. 307 4- 140 
= 2 + . Place 2 in the root. Add 2 to 140. Mul¬ 

140 

2 

3 07 

tiply 142 by 2. Subtract. Annex 06. 

142 

2 84 

3. Trial divisor is 2 X 72. Annex 0. 2306 4- 
1440 = 1+. Place 1 in the root. Add 1 to 1440. 

1440 

1 

23 06 

Multiply 1441 by 1. Subtract. Annex 56. 

1441 

14 41 

4. Trial divisor is 2 X 721. Annex 0. 86556 4- 
14420 fl 6 + . Place 6 in the root. Add 6 to 14420. 

14420 

6 

8 65 56 

Multiply 14426 by 6. No remainder. 

Note. — The root, correct to tenths , is 72.2. 

14426 

8 65 56 












230 


ALGEBRA 


EXERCISE 125 

Find the principal square root of: 


1 . 

1296 

6. 17,689 

11. 164,025 

16. 

9.4249 

2. 

1849 

7. 33,489 

12. 1474.56 

17. 

761.76 

3. 

3721 

8. 82,369 

13. 19.1844 

18. 

5913.61 

4. 

7396 

9. 96,100 

14. 4160.25 

19. 

8.5264 

6 . 

8836 

10. 43,264 

15. 62.8849 

20. 

64.6416 


164. Approximate values of numbers. The number -J- cannot 
be expressed exactly as a decimal. We know that ^ = .333 (etc.). 
.3, .33, .333, etc. are all approximations to -J-. 

.3 is the value of correct to tenths. 

.33 is the value of ^ correct to hundredths. 

.333 is the value of ^ correct to thousandths. 

Only approximations to the square roots of many numbers 
can be found. 


165. Finding the approximate square root of a number. 

Example. Find V2 to three decimal places. 


Solution. 

20 + 4, or 
280 + 1, or 
2820 + 4, or 


1. 4 1 4 



2.00 

00 

00 


1 





1.00 



24 

96 





4 

00 


281 

2 

81 




1 

19 

00 

2824 

1 

12 

96 


Therefore the square root of 
2 to three decimal places is 1.414. 
The root, * correct to hundredths, 
is 1.41, since the figure in the 
third decimal place is less than 
5. If it were more than 5, the 
figure in the hundredths’ place 
would be increased by 1. Thus, 
3.146, correct to hundredths, is 3.15. 


EXERCISE 126 

Find the square root correct to hundredths : 


1 . 

3 

4. 

7 

7. 

13 

10. 

17 

13. 138 

16. 

526 

2. 

5 

5. 

10 

8. 

14 

11. 

19 

14. 249 

17. 

1182 

3. 

6 

6. 

11 

9. 

15 

12. 

21 

15. 307 

18. 

2067 







SQUARE ROOT AND RADICALS 


231 


166. A table of square roots of numbers from 1 to 100 is 
given on page 331. These can be used to find the square roots 
of other integers, and of fractions. 

Example. Find \Zl80 correct to tenths. 

Solution. Vl80 = V3(h5 = 6 V 5 = 6 X 2.236 = 13.416, or 13.4. 

167. The square root of a fraction. 

Example 1. Find a/j correct to hundredths. 

Method a. § = 1.5. The square root can be found as in § 168. 
It cannot be got mentally. 



Method b. 


! , correct to hundredths, is 1.22. 


Method b is clearly the one to use, since only one square root 
needs to be computed or secured from the table. 

Rule. — To find the square root of a fraction: 

1. Change the fraction to an equivalent fraction whose de¬ 
nominator is the smallest possible perfect square. 

2. The square root of the fraction equals the square root of 
its numerator divided by the square root of its denominator. 

3. Express the result of Step 2 in simplest radical form, and 
also in decimal form, to three places. 

Example 2. Find correct to hundredths. 

Solution 1. The smallest perfect square into which 8 can be changed 
is 16. Multiply both terms of the fraction by 2. 


2 . 



3. 


4 


I-, correct to hundredths, is .61. 
8 


[Examples are to be found on page 232.] 




232 


ALGEBRA 


EXERCISE 127 


Find the square root correct to hundredths: 


1. 

2 

■3 

4. 

2 

9 

7. 

2 

T 

10. 

7 

¥ 

13. 

4 

XT 

16. 

"V 8- 

2. 

3 

5 

5. 

3 

4 

8. 

f 

11. 

3 

7 

14. 

5 

1 2 

17. 

1 1 

3. 

5 

"8 

6. 

5 

6 

9. 

9 

¥ 

12. 

3% 

15. 

9 

T3 

18. 

¥ 


Miscellaneous Examples 

19. By the formula h = Vc 2 — t 2 , find h : 

a. when c = 35 and t = 10. 

b. when c = 43 and t = 18. 

c. when c = 13^- and t = 5^-. 

20. If V = a/2 gd, find V when g = 32.16 and d = 25. 


168. The approximate value of a mixed expression. 

Example. Simplify f — V 


Solution 1. 

2 . 

3. 


3 _ 12 = 3 _ [i 4 

7 \7 7 \49 

= 3 _ vn 

7 7 

_ 3 —Vl4 
7 


Find the value of the expression correct to hundredths. 

Solution. 3 -y T? = 3 - 3 741 = = - .105, or - .11. 

7 7 7 


EXERCISE 128 

a. Simplify the following expressions; 
correct to hundredths: 


6 +Vl2 

- 5 +VT8 
4 - V20 

- 3 -Vs 

+ 7+V24 


6 . 

7. 

8 . 

9. 

10 . 


i-v'* 

3. -l-Vi 
4 v 8 

t+v^ 


6. express the results 








SQUARE ROOT AND RADICALS 


233 


169. A radical is an indicated root of a number. The square 
root may be indicated either by the radical sign, as V2, or 
Va + b ; or by \ used as an exponent, as 4 ^, or x*. 

Hence, 4* = 2; (9 x : 2 )* = 3 x; (x 2 — 2 xy + y 2 )^ = x — y. 

170. A quadratic surd is the indicated square root of a num¬ 
ber which is not a perfect square; as V3, or 'Vox, or V 3 m. 
The principal square root is implied. 

A quadratic surd is in its simplest form when the number 
under the radical sign is an integer or integral expression which 
does not have any perfect square factors. • 

Thus, 5(24 z 3 )* = 5(4 x 2 • 6 xfi = 5(4 x*)% • (6 »)* = 5 • 2 x • or 
10 xV 6 x. 

Also, V £ (a — z) 2 t/ = V f (a — x) 2 • y = -~ x - Vq y. 

O 

EXERCISE 129 


Simplify the following radicals : 


1. 

Vs 

7. 

V72 a 4 13. 

3Vi96 

19. -jV 144 y 3 

2. 

V50 

8. 

(49 a: 6 )* 14- 

5(100 to)* 

20. -(16z 7 )* 

3. 

(12)* 

9. 

Vf? IB. 

4V121 y* 

2 

21. yV 24 x 2 

4. 

(75)* 

10. 

(81ft 3 )* 16. 

i(98b 2 )i 

22. mV 169 m 3 

5. 

(18)* 

11. 

(24 a 4 )* 17. 

|(4-91)* 

23. ?i(40 m 5 ) ¥ 

6. 

V36 a 3 

12. 

V20x 2 18. 

|V25-3 ri- 

1 24. V2c • 14 c 

25. 

(1000 l 6 )* 


31. (108 l 4 )* 

37. 

V5 c(a — 6) 2 

26. 

(120 a 3 )* 


32. V200 rs 3 

38. 

[(* - 4) 2 ]* 

27. 

V84 cd 2 


33. V250 .s' 5 

39. 

Va(y - 5) 2 

28. 

V156 m 7 


34. (175 cd)* 

40. [m(t - 6) 2 ]* 

29. 

(96 a 3 6)* 


36. V125 m 8 

41. 

V 5a 2 (35—5a) 2 

30. 

(28 !/Z 2 )* 


36. (180 r 7 )* 

42. 

Vl2(8x— 7i/) 2 























234 


ALGEBRA 


171. Simplifying square roots of fractional expressions. 

Example 1. ^ a 5 = a? ■ a = = 9 lV2Ui. 

Example 2. 3(24*)* -■ S (lf-f = 

= 3 ' 11 V5, or — y/b. 

5 5 

Example 3. = ^ 1# • a* ^ • a • 11 


EXERCISE 130 

Simplify the following radicals: 


1. 

2. 

v'f 

(2b\h 

V36V 

12. 

J3a 
*4 b 

25. 

26. 

V3f 

(2# 

37. 

(¥)* 


_ 

13. 

30(1)* 



38. 

/4 a; 2 \^ 

3. 

a/ 81 

V T2T 

14. 

14Vf 

27. 

V9f 

(V 

4. 

/'144\i 
\ 1 6 9 / 

15. 

16V& 

28. 

(9# 

39. 

/25^ 

6. 

Vi 


. v 1 

29. 

V5f 


\ 49 ^3 


3 

16. 

21(f)* 





6. 

7. 

a)* 

17. 

18. 

11(A)* 
20 Va 

30. 

31. 

(8A)* 

V8f 

, v 1 

40. 

^64 m 4 ^ 2 

/I 21 7/ 6 \^ 

8. 

vl 

19. 

9(ifi)* 

32. 

(5A)* 

41. 

/ y \ 

\ 7 / 

9. 

i 

20. 

21. 

7\/121 
* v IT 

V5f 

33. 

34. 

viof 

(3|)* 

42. 

/si z 10 

* li 

10. 

(¥) 

22. 

(4# 

35. 

Vi 

43. 

Vt 8 

11. 


23. 

24. 

Vsf 

(7f) 4 

36. 

v* 

44. 

o 

/II w%V 
\ 16 ) 

















SQUARE ROOT AND RADICALS 


235 


172. A surd expression is an expression involving one or more 
surds. 

A surd expression is simplified by first simplifying the surds, 
and then performing all other indicated operations. 

Similar surds are surd expressions whose surd factors are the 
same; thus, 3^2 and — 7 V 2 are similar surds. 


173. Addition and subtraction of quadratic surds. 

Example 1. Simplify V20 + a/45. 

Solution 1. V20 +V45 =V4 • 5 +V9 • 5 

2. = 2V5 + 3V5, or 5V5. 


Example 2. 
Solution 1. 

2 . 

3. 


Simplify V J + VJ. 



3V2 , V2 

4 + 2 

3 V 2 + 2V2 Qr 5V2 


EXERCISE 131 


Find the simplest radical form and also the decimal value to 
three places of the following expressions : 


1. V12+V27 

11 . Vf-Vj 

2. V75-V48 

12. f-V* 

3. Vl8^+\/32 

13 . V28 — 3V7 - V63 

4. V45-V5 

14. V50-V18+V8 

5. 3A/6-V24 

15. V80 — V45+V5 

'■> 

CO 

100 

<D 

16. V48 - 2V3+V300 

7. V%+V8 

17. V200 — V98 — V72 

8. Vf+Vf 

18. 3V^ — |V3 — 2V75 

9. 

19. 9Vf -(12) 1 * + 6Vf 

10 . V^+ViO 

20. V 18a 2 + aV 8 — 2V20 2 


Note. — Additional examples can be found on page 291. 



236 


ALGEBRA 


174. Multiplication and division of quadratic surds. 

Rule.— The product of the square roots of two positive 
numbers equals the square root of the product of the numbers. 

Example. Find the simplest radical form and also the 
decimal value of 2 V 3 X 5a/ 2 correct to hundredths. 

Solution. 1. 2V3 X 5^2 = 10V3 • 2, or 10^6. 

2. 10V6 = 10 X 2.449, or 24.49. 

EXERCISE 132 

Find the simplest radical form and the value to hundredths: 


1 . 

V2 

Vi 

8. 

Ve- 

Vl2 

15. 

3 V 2 -4V5 

2. 

v§ 

■s /12 

9. 

V 7 • 

V21 

16. 

2 V 6 • 5 V 3 

3. 

Vs 

Vl5 

10. 

Vn 

• V22 

17. 

6V20 • 3VlO 

4. 

V 2 

Vl8 

11. 

V 7 • 

Vs 

18. 

2 V 5 • 3Vl5 

5. 

V~2 

V50 

12. 

Vl4 

• V 21 

19. 

Vf.Vi 

6. 

Vs 

V20 

13. 

V 9 • 

Vl8 

20. 

Vf-Vf 

7. 

V 3 

Vs 

14. 

Vl2 

■ Vl5 

21. 

3Vf • 2V^ 


Rule. — The quotient of the square roots of two positive 
numbers is the square root of the quotient of the numbers. 

Example 1. V5 -h V15 = = v / i. But Vj = V| = i V3. 

EXERCISE 133 

Find the following quotients : 


1. 

VTi - 7 -V 2 

8. V51 -rV 17 

15. 

4 V 45 - 4 - 3 V 5 

2. 

V 50 +V2 

9. V 32 - 4 -V 4 

16. 

11 V 99 4 - 3 V 1 T 

3. 

V 98 -fVS 

10. VlQ2 4- 3V2 

17. 

v*w| 

4. 

V 20 +V5 

11. V363 -4- llV3 

18. 

Vf -4-Vf 

5. 

Vl2 -Vs 

12. Vl28 -f- 3^2 

19. 

VtV 2 +V 9 I? 

6. 

5 V 2 I + V 3 

13. 3V24 +V6 

20. 

V49 m 2 

7. 

6^75 -i- 2 V 3 

14. 7V35 5V7 

21. 

V4x* -7-V95 




SQUARE ROOT AND RADICALS 


237 


175. Rationalizing the denominator. Positive and negative 
integers or fractions are rational numbers ; a surd is an irrational 
number. The following examples illustrate another method 
of dividing a number by a radical. 

5 = 5 • V2 = 5V2 

V2 V2 • V2 2 

20 = 20 • V3 = 20V3 = 20V3 = 10V3 . 

Vl2 Vl2 * V3 V36 6 3 


Example 1. 
Example 2. 


In both examples, the numerator and denominator of the 
fraction are multiplied by a surd which makes the new denomi¬ 
nator the square root of the smallest possible perfect square. 
This process is called rationalizing the denominator of the fraction, 
because the denominator of the result is a rational number. 

EXERCISE 134 


Rationalize the denominator: 

(In Exs. 1-10 find results to nearest tenth.) 


1 

1 

Q 

6 

1 K 

13 

00 

b 

1* 

Vs 

O. 

a/18 

10. 

Vis 


V^a 

2. 

3 

9. 

15 

16. 

28 

23. 

me 


VS 


V20 


Vli 


Vi 

3. 

7 

10. 

7 

17. 

5 

24. 

qVx 


V2 

Vn 

V45 

Wy 

4. 

5 

11. 

6 

18. 

7 

25. 

15VX 3 


V§ 


vlo 

V98 

V3^ 

5. 

9 

12. 

20 

19. 

15 

26. 

20V2xf 


Vs 


V50 

a/75 

VSy 

6. 

8 

13. 

15 

20. 

20 

27. 

10V6 xy 3 


Vl2 


V30 

V24 

VlO xy 


8 

14. 

2 

21. 

2 a 

28. 

25Wt 

7. 

V20 

V2 

Va 

3V5r 














238 


ALGEBRA 


176.* Multiplication of binomial quadratic surds. 

Note. — This topic is optional. It is (probably) not implied as a 
requirement by the C.E.E.B. 

Example 1. Multiply 2 + 3V2 by 2 — 3^2. 

Solution. (2 + 3V2) (2-3 V2) = (2) 2 - (3 V2 ) 2 

= 4 - 18, or - 14. 

Example 2. Find (3 — 2 V 5 ) 2 . 

Solution. (3 - 2V5) 2 =(3) 2 - 2(3)(2v / 5) + (2\ / 5) 2 
= 9 - 12V5_+ 20 
= 29 - 12V5. 


EXERCISE 135* 


Perform the following indicated operation : 

1 . 

(2 -Vi) (2-Vi) 

13. 

(5 - Vi) (5 +Vi) 

2 . 

(5 +vD)(5 +V6) 

14. 

(5 - 2Vi) 2 

3. 

(3 -V 2) 2 

16. 

(5 - 2 Vi) (5 + 2 Vi) 

4. 

(9 - V 5) 2 

16. 

(4 - 2 V 3 ) (4 - 2 Vi) 

5. 

(- 2 +V 3) 2 

17. 

(4 + 2Vi) 2 

6 . 

(- 2 +V3)(— 2 — Vi) 

18. 

(6 - 2 V 5) 2 

7. 

(3 -V5)(3 +V5) 

19. 

(5 + 2 V 6) 2 

8 . 

(4 - Vi) 2 

20 . 

(6 - 3V5)(6 + 3 V 5 ) 

9. 

(3 - 2V2)(3 + 2 V 2 ) 

21 . 

(- 2 + 3 V 5) 2 

10 . 

(3 - 2 V 2) 2 

22. 

(- 2 - 4Vi) 2 

11 . 

(2 + 3 V 2 ) (2 + 3 V 2 ) 

23. 

(- 2 — V6)(— 2+V6) 

12 . 

(3 - 2V5)(3 + 2 V 5 ) 

24. 

(V6 - 2 V 5) 2 

26. 

(2 + Vi) 2 - 2(2 + Vi)- 

-5 


26. 

(3 - V 5) 2 + 3(3 - V5)- 

-4 


27. 

(- 2 + V6) 2 + 4(—2 + V6) — 

•2 

28. 

(- 2 - V6) 2 + 4(- 2 - 

V6)- 

- 2 


Note. — Exercise 168, page 291, can be done now. 


SQUARE ROOT AND RADICALS 


239 


177.# Simplifying cube roots. 

Note. — This topic is not required by the Regents. It is part of 
the requirements of the C.E.E.B. 

Later on in mathematics you will learn how to prove that every 
number has three cube roots. One root is called the principal 
root and is to be given when nothing is said to the contrary. 

The cube root of a is indicated either by or by a?. If a 
is positive, the principal cube root is positive; thus \^8 = 2, since 
2 • 2 • 2 = 8. If a is negative, the principal cube root is negative; 
thus (- 8)* or \^8 = - 2, since (- 2)(- 2)(- 2)= - 8. 

Example 1. v 7 — 64 a 3 = p — 4) 3 a 3 = — 4 a. 

Example 2. (125 xy 4 )' 3 ' = (5 hfxy)^ = 5 y{xy)* or 5 y^xy. 

Example 3. '■ - = = ~ 


EXERCISE 136 


Simplify the following radicals : 


1 . ^8 

2. (27)* 

3. V- 64 

4. (- 125)* 

6. </- 27 

6 - (-# 

7. (- 81)* 

8. a/24 


9. </- 40 

10. (54)* 

11 . Vi 



13. 

V 7 250 a 

24. 

y-JrV 6 

32. 

a/ 

_ 2 

¥ 

14. 

(-128 a; 2 )* 

26. 

(rf# 

33. 

a! 

^ c 

16. 

16. 

V— 54 x? 
(72 to 6 )* 

26. 

* 64 

34. 

* ab 

/mV 

\tfy) 

17. 

ZV- 8 m 

27. 

(?) 

36. 


18. 

4(— 27 <)*• 

28. 

(— J=-m 3 w) T 

36. 

(- 

■*)* 

19. 

- 2v / 64c 6 



37. 

V(a — 6) : 

20. 

4(^V z 5 )* 

. 1 

29. 

yliz 

38. 

3 

/2 m 

W 2 

to to 
to H*. 

(-125 m)* 

V- 81 2/ 4 

30. 

^ r 

39. 

3 

p 

2 a 2 

23. 

3 /m 

31. 

/ cr 2 V 

\2 y/ 

40. 

1. 

a 

p 

’54 


















240 


ALGEBRA 


REVIEW EXERCISE XI 

c , p m mx + b 7 x — c 

1 . Solve for x : a. T -- b. c =--• 

n x + 1 

2. Solve: - f) - |(i/ + -J) + i = 0. Check. 

3. Solve for x : -J(3.5 x — 7 t) = -J( 1.8 x + 3 t). Check. 

4. Solve — = —: a. for C; b. for R; o. for r. 

c r 

5. Separate 84 into three parts which are to each other as 
3:4:5. 

6 . One of two furnaces in a building will burn 10 tons of 
coal in 14 days; the other, 10 tons in 20 days. How long will 
10 tons last if both furnaces are run at the same time ? 

7. A motorcycle policeman chased a speeding motorist who 
was \ mile away. If the motorcycle traveled 55 miles per hour, 
and overtook the car in 5 minutes, what was the speed of the car ? 

W 

8. a. Solve the formula D = --for W. 

W — w 

b. Find W if D = 1.5 and w = 25. 

c. Find W, correct to tenths, if D = 1.6 and w = 25. 

9 . Simplify _ 3*_ +2 y _ 2 x-Jjy_ Check . 

3 9 6 

let x = 2 ; y = 1 . 

10. 3 y - 4 - L (7 y - 9 ) = |(6 + 2-=^). Check. 

11 . The denominator of a certain fraction is 2 more than 
the numerator. If 7 be added to both the numerator and the 
denominator, the value of the resulting fraction is Find 
the fraction. 

12 . Solve: |(x — l)(as — 2) — -*)=£*(* + 2). 

13. The sum of the squares of certain three consecutive 
even integers is 596. Find the integers. 

14. Simplify: 2 V 25 y — 7V4 y+ lW§y. 









SQUARE ROOT AND RADICALS 


241 


REVIEW EXERCISE XII 

1. Add cy 2 — dy + e and my 2 + ny — /. 

2. Simplify 3.8 a - [2.5 a - 2j.6 a - 3.1 j - 7.4]. 

3. a. By the formula S = P( 1 + r) 2 , find 8 when P = 
$1500 and r = 5%. b. Solve the same formula for P. 

4. Factor: a. C 2 — 4 CD - B 2 + 4 D 2 . 6. 81 z 2 ” — 16 y 2n . 

5. Simplify the following expression. Check. 

( V + 4 _ 4 \ ^ / y - 5 3 \ 

^2/ — 2 V ~ 3/ \2/ — 3 y — 1/ 

6. a. Express as a formula: the dividend (D) equals the 
divisor (d), times the quotient (g), plus the remainder (P). 

6. By this formula, find the dividend when the divisor is 
13, the quotient is 11, and the remainder is 8. 

7. a. Does -| satisfy the equation 5 x — 3 = 5 — 7z? 
b. Is -J a root of the equation 3 x 2 + x = 2 ? 

8. Solve to the nearest tenth : 

2.6 x - 3.5 = 14.8 - 3.2 x 

9. a. Represent by a curved line graph the relation A = s 2 . 
(Make a table of values of A when s = 0, 1, 2, . . . etc. to 10; 
use the horizontal axis for values of s.) 

b. Find, on this graph, A when s = 2.5; when s = 6.5. 

c. Find s when A = 40; when A = 75. 

10 . a. Solve the formula v = —— for M. 

M + m 

b. Find M when V = 15, m = 25, and v = 10. 

11. Solve: -jt(8 — t) + -J t — 1-J = %(t + 6). Check. 

12. Represent by a broken line graph : 


Average daily 

TEMPERATURE 

- 10° 

1 

Ol 

o 

O 

O 

5° 

10 ° 

15° 

to 

o 

o 

Coal consumed in 

FURNACE IN DAY 

400 lb. 

350 lb. 

300 lb. 

270 lb. 

240 lb. 

200 lb. 

150 lb. 

















XIV. QUADRATIC EQUATIONS 


178. An incomplete quadratic equation is an equation in one 
unknown in which the unknown does not appear in the denomi¬ 
nator of a fraction, and in which only the second power of the 
unknown does appear; as, 5 x 2 — 37 = 0. 


179. An incomplete quadratic equation has two roots. 


Example 1. x 2 = 16. 

Solution. Taking the square root of both sides, ± x = ±4. 

This means: (1) + x = + 4 (3) — x = + 4 

(2) + x = - 4 (4) — x = — 4 

However, if both members of equation (3) are multiplied by — 1, 

we get + x = — 4, which is equation (2) ; and if both members of 
equation (4) are multiplied by — 1, we get + x = +4, which is 
equation (1). We actually get only two equations, and not four. 

We get these two equations if we use only the positive root on the 
left side and the two roots on the right side; that is, + x = ± 4. 

Example 2. Solve the equation + — = — + — • 

3 m 12 m 


Solution. 1. 

2. Mi2m 

3. Simplifying, 

4. M 7 

5. m = 


2 m i 3 _ m , 12 
3 m 12 m 
8 m 2 + 36 = m 2 + 144. 

7 m 2 = 108. 
ra 2 = 

± vToi = ± 6 Vf = ± f- V21. 


6. m = ± | • (4.582) = ± = ± 3.927. 


Rule. — To solve an incomplete quadratic equation : 

1. Simplify the equation until it takes the form x 2 = a number. 

2. Take the square root of both members of the equation, 
placing the + sign in front of x, and the double sign (±) before 
the principal square root of the right member. 

242 



QUADRATIC EQUATIONS 


243 


EXERCISE 137 


Find the roots: 


1 . 11 x 2 - 99 = 0 

2 . 5 x 2 — 125 = 3 x 2 - 27 

3. 13 c 2 + 8 = 88 - 7 c 2 

4. (y + 5 ) 2 = 10 y + 34 

5. (2 + 3 ) 2 - 6(2 + 3)= 7 

6. 4 x 2 = 9 

7. 9 a ; 2 = 2.25 

8. .6 x = 8.64 

9. 2 a : 2 = 4J. 



25 

x 


11 . 



4 

x + 1 


12 3 _i_ 3 

X- 3 x+3 


17 

4 


13. —- —5— = —2 
x + 1 X — 1 

14 _ 9 __ — 3 

(x + 1) (x - 1) 


15. 3(2x - 1)+ 2 x(x + 2) = 10 x + 29 

16 4 a/ 2 + 5 _ 3 a/ 2 + 1 _ 5 y 2 15 
3 6 9 


17. 

18. 

19. 

20 . 
21 . 

22 . 

23. 

24. 

25. 

26. 

27. 

28. 
29. 


{a + x) (b — x) + (a — x) (b + x) = 0 
(a — x)(b — x)= (1 — ax)(l — bx) 
Solve a 2 + x 2 = m 2 , for x. 

Solve cx 2 — d, for x. 

Solve s = ^ gt 2 , for t. 


Solve / = rr ^, for v. 
r 

Solve v = 7r r 2 h, for r. 

Solve 5 = 4 7r r 2 , for r. 

Solve a 2 + b 2 = c 2 , for a. 

Solve a 2 + b 2 = c 2 , for b. 

Solve a 2 + b 2 = c 2 , for c. 

Solve for x: (x + 4) 2 — 4(x + 4) = 12. 


Solve for x: 


x 2 + x + 1 _ x 2 + 5 x + 7 

X — 1 X + 1 














244 


ALGEBRA 



180. A right triangle is a triangle which has a 
right angle for one of its angles; the side opposite 
the right angle is the hypotenuse (side AC) ; the 
side BC is the base and AB is the altitude. 


In a right triangle , the square of the hypotenuse equals the sum 
of the squares of the other two sides; or b 2 = a 2 + c 2 . 


EXERCISE 138 

Find the remaining side of the right triangle, correct to 
hundredths, if two sides are: 


1. 

a = 5; 

b = 12 

4. 

a = 30; 

b = 72 

2. 

a = 10 

; b = 24 

5. 

c = 39; 

b = 55 

3. 

a = 8; 

c = 10 

6. 

c = 51; a = 24 


7 . Find the diagonals of the rectangle whose sides are 50 
and 120. 

8. Find the base of the rectangle whose altitude is 18 and 
whose diagonal is 30. 

9. A ladder whose foot stands 9 feet from a side of a building 
just reaches a window 40 feet from the ground. How long is the 
ladder ? 

10. When one side of a rectangle is 30 feet and the diagonal is 

34 feet, what is the other side ? 

11. Find the area of the rectangle whose base is 32 feet and 
whose diagonal is 60 feet. 

12. The foot of a ladder 51 feet long is 24 feet from a side of a 
building. How high on the side of the building does it reach ? 

13. Find the base of the right triangle whose hypotenuse is 

35 feet and whose altitude is 20 feet. 

14. If the diagonal of a rectangle is 65 inches and the base 
of it is 3 times the altitude, what are the dimensions of the 
rectangle? (Draw a figure.) 




QUADRATIC EQUATIONS 


245 


15. a. The altitude of a rectangle is a inches. The base is 
3 times as long. What is the length of the diagonal ? 

b. Using the result of part a as a formula, how long is the 
diagonal when a = 5 ? a = 8 ? 

16. Solve the formula b 2 = a 2 -f- c 2 : a. for b ; b. for c. 

17. Triangle ABC is an isosceles triangle, 
since AB = AC. AD is the altitude. 

a. Compare BD and DC by measurement. 

b. Suppose AB = 15 inches and BC is 20 
inches. Find AD. 

c. If the equal sides of an isosceles triangle are each 25 inches 
long, and the altitude is 15 inches, find the base. 

181. A complete quadratic equation is an equation having 
only one unknown, in which the unknown does not appear in 
the denominator of the equation, and in which the first and 
second powers of the unknown do appear. 

Thus, 2 z 2 — 3x — 5 = 0 is a complete quadratic equation. 

182. * Solution of a complete quadratic equation by the 
factoring method was taught on pages 134 to 135. Review § 105. 

Note. — Recall that this topic is optional but that it is required by 
the C.E.E.B. for cases such as examples 1-10 below. 



EXERCISE 139 

Solve the following quadratic equations by factoring: 


1 . x 2 - 7 x + 12 = 0 

2. y 2 — 5 y — 14 = 0 

3. t 2 + 8 t + 15 = 0 

4. a 2 + 5 a = 6 
6 . r 2 = 3 r 10 

6 . m 2 + 4 m = 45 

7. w 2 = 13 w - 22 

8 . m 2 = 144 


9. z 2 = 12 - 11 z 

10. a 2 + 11 a = 42 

11. 2x 2 + x-6 = 0 

12 . 3 m 2 + 11 m, = 20 

13. 4 y 2 + 17 y — 15 = 0 

14. 3 t 2 = 8 t + 16 

15. x 2 -^x-l=0 

16. x 2 + j irx = 2 




246 


ALGEBRA 


183.* Solution of complete quadratics by completing the 
square. 

Preparation. 1. Find : a. (x + 4) 2 ; b. (y — 5) 2 ; 

c. (* + i) 2 ; d. (w- i) 2 ; e. (r- f) 2 ; f.(s- f ) 2 . 

2. Make a perfect square trinomial of each of the following 
expressions (see page 257), and give its square root. 

a. x 2 + 8 x; b. y 2 — 14 y; c. w 2 — 20 w; d. m 2 — m. 

3. Notice that the coefficient of the second power in each of 
these expressions is 1; that the term added in each case is the 
square of one half the coefficient of the first power; thus, in 
o, you add (f ) 2 . 

This process is called completing the square. It is used in 
solving quadratics. 

Example 1. Solve x 2 — 12 x + 20 = 0. 

Solution. 1. S20 x 2 — 12 x = — 20. 

2. We can complete the square of the left member by adding (V-) 2 
or 36. But then we must also add 36 to the right member. 


3. A3fl 

z 2 - 12 x + 36 = - 20 + 36. 


4. 

(x - 6) 2 = 16. 


5. V 

x — 6 = =•= 4. Read Rule, page 242. 

6. 

.*. x — 6 = + 4, or x = 10. One root. The other 

7. 

.*. x — 6 = — 4, or x = 2. root. 

Check. 

These roots check when substituted in equation 1. 

Example 2. Solve x 2 — 6 x — 5 = 0. 


Solution. 

1. Ag x 2 — 6 x = 5. 


2. 

(-6) -5-2= -3; (— 3) 2 = +9. 

3. A, 

x 2 -6x + 9 = 5 + 9. 


4. 

(x - 3) 2 = 14. 


5. 

x — 3 = =*= Vl4. 


6. When x - 3 = + V4i, x = 3 + Vl4. 1 

1 These are the roots 

When x — 3 = — Vl4, x = 3 — Vl4. J 

' in radical form. 

7. 

x = 3 + Vl4 = 3 + 3.741 = 6.741. 1 

These are the roots 


x = 3 - Vl4 = 3 - 3.741 = -.741. J 

in decimal form. 


QUADRATIC EQUATIONS 


247 


Rule. — To solve a quadratic equation by completing the 
square: 

1. Simplify the equation; transpose all terms containing the 
unknown number to the left member, and all other terms to 
the right member so that the equation takes the form 

ax 2 + bx = c. 

2. If the coefficient of x 2 is not 1, divide both members of the 
equation by it, so that the equation takes the form 

x 2 -f px = q. 

3. Find one half of the coefficient of x; square the result; 
add the square to both members of the equation obtained in 
Step 2. This makes the left member a perfect square. 

4. Write the left member as the square of a binomial; ex¬ 
press the right member in its simplest form. 

5. Take the square root of both members, writing the double 
sign, ±, before the square root in the right member. 

6. Set the left square root equal to the + root in the right 
member of the equation in Step 5. Solve for the unknown. 
This gives one root. 

7. Repeat the process, using the — root in Step 5. This 
gives the second root of the equation. 

8. Express the roots first in simplest radical form, and then 
in simplest decimal form. 

9. Checks, a. The decimal values usually do not exactly sat¬ 
isfy the equation since they are usually only approximate values 
of the roots. Only the radical values exactly satisfy the equation. 

b. A second method illustrated by Example 2, page 246. 

Adding the roots: 6.741 + (— .741) = + 6. This is the 
negative of the coefficient of x in the original equation. 

Multiplying the roots: (6.741) X (- .741) = - 4.995, which is 
almost - 5. This is the term free from x in the original equation. 

This second method of checking may be used only when the 
coefficient of x 2 is 1. 


248 


ALGEBRA 


EXERCISE 140 * 

Solve by completing the square. Express surd roots correct 
to tenths. Check as directed by your teacher. 

1 . x 2 — 8 a; + 15 = 0 5. y 2 — 4 y + 1 = 0 

2 . y 2 + 8 y + 12 = 0 6. z 2 = 6 z - 4 

3. z 2 + 2 z - 35 = 0 7. w 2 + 10 w + 22 = 0 

4. x 2 — 1 x — 2 = i) 8. s 2 — 8 5 + 13 = 0 

9. Solve the equation x 2 — 9 x + 7 = 0. 


Solution. 

1. S 7 a: 2 -9x=-7. 

2. 

(-9)+2 =-f; (-f) 2 

3. A^ 

x 2 -9x + ^ = Y~ 7 - 

4. 

(*-f) 2 =¥• 

5. 


6. 

„ 9 , a/ 53 9 +V53 

2 2 2 

7. Then 

9 +V53 9 + 7.280 16.28 

1 2 2 2 


and 


8.14, or 8.1 


. 9 ~ V53 = 9 - 7 280 = 1^2 _ g6 r 9 
2 2 2 


10. x 2 + 3 x — 4 = 0 

11 . z 2 — 3 23 “f* 2 = 0 

12 . y 2 + 5 y = 14 

13. m 2 + m = 6 

14. c 2 - 7 c + 12 = 0 

15. r 2 - 9 r = 10 

16. t 2 - 7 t + 6 = 0 

17. a 2 — 15 a + 56 = 0 

26. Solve the equation x 2 + f x = 
Solution. 1. x 2 + f x = 


18 . s 2 - 3 5 + 1 = 0 

19. x 2 — x — 5 = 0 

20. m 2 = 5 m — 5 

21. t 2 + 7t + 11 = 0 

22. p 2 — 9 p + 19 = 0 

23. w 2 = 1 — w 

24. x 2 + 3 x — 1 = 0 

25. y 2 — 2 = 5 y 


2 . 

3. 

4. 


t of # = i 


(£) 2 = i 


* 2 + tz + ^-= t + 

(* + i) 2 = H- 

(Complete the solution.) 







QUADRATIC EQUATIONS 


249 


27. x 2 + %x~i = 0 

28 . + i = 0 

29. y*-iy+t = 0 

30. a; 2 -fa;-2 = 0 

31. x 2 - $x - f = 0 

32. z 2 - l°- 2 - s = o 


33. x 2 + %x = 3 
(Hint. — i of ^ = ^.) 

34. x 2 - | X = i 

35 . y 2 -iy-2\ = o 

36. a; 2 ~ f * - £ = 0 

37. 2 2 -| z + i = 0 


38. Solve the equation 3 a; 2 — 5 a; + 2 = 0. 
Solution. 1. 3 x 2 — 5 x + 2 = 0 . 

2. In order to make the coefficient of a: 2 be 1, 

D 3 x 2 — | x + f = 0. 

(Complete the solution.) 


39. 2 a; 2 + 5 x + 2 = 0 

40. 3 a; 2 — 10 £ + 3 = 0 

41. 6 x 2 — x — 2 = 0 

42. 4 p 2 — 4 p + 1 = 0 

43. 5 r 2 = 2 r + 7 

44. 3m 2 — 4m — 4 = 0 


46. 7 a; 2 - 2 x - 3 = 0 

46. 8 y 2 — 4 y — 1 = 0 

47. 9 w 2 — 6 w — 4 = 0 

48. 6 t 2 - 8 t — 5 = 0 

49. 12 s 2 - 10 s - 9 = 0 
60. 10 w 2 — 4 w — 3 = 0 


184.* Solution of literal quadratics by completing the square. 


Example. Solve for x the equation x 2 + 3 ax — 4 a 2 = 0. 
Solution. 1. x 2 + 3 ax = 4 a 2 . 

/3a V _ 9 a 2 

\2 ) 4 ' 

9 a 2 

T/ 4 


* of3a = (¥> 

3 ax + ^ 

(•+¥)’ 


2 . 


3. A^9a2^ x 2 + 3 ax + 


4. 

5 . 


= — + 4 a 2 . 
4 


. 9 a 2 + 16 a 2 


•••* + T- ± 


4 

/25 a 2 . 


a . „ 3 a i 5 a 

6 . . 


2 a 


or a. 


3 a _ 5_a 
2 2 
8 a 


, or — 4 a. 




250 


ALGEBRA 


EXERCISE 141* 


Solve for x by completing the square : 


1. x 2 + 2 ax — 3 a 2 = 0 

2. x 2 + 4 bx = 12 b 2 

3. x 2 — 6 bx + 5 b 2 = 0 

4. 4 x 2 + 4 xt + t 2 = 0 

5. x 2 + 2az-b3 = 0 

6. x 2 + 4 ax -f- b = 0 

7. x 2 — 6 bx — c = 0 


8. £ 2 — 2 mx = 1 — 2 m 

9. z 2 -f- 6 x = t 2 + 6 t 

10 . x 2 — 4 ax = b 2 — 4 ab 

11 . a;r 2 + 4 £ + 1 = 0 

12. ax 2 + x + 1 =0 

13. ax 2 + 2 dx + m = 0 

14. Ax 2 + Bx + C = 0 


185. Solution of quadratic equations by a formula. All 

quadratic equations having one unknown may be put in the 
form ax 2 + bx + c = 0. 

2 x 2 — 3 x — 5=0 has this form, a = 2, b = — 3, and c = — 5. 

If we solve ax 2 + bx + c = 0 for x, the roots may be used 
as formulas for the roots of any quadratic equation. 

Solution. 1. ax 2 + bx + c = 0. 

2. D a z 2 + h - • x + - = 0. 

a a 


3. Sc x 2 + — • x = — -• 

a O/O/ 



— fr + Vb 2 — 4 ac 
2 a 


Note. — Not required either by the Regents or by the C.E.E.B. 










QUADRATIC EQUATIONS 


251 


Example showing the use of this important formula. 

Solve the equation 2 x 2 — 3 x — 5 = 0, by the formula. 
Solution. 1 . Comparing this equation with ax 2 + bx + c = 0 
a = 2, b = — 3, c = — 5. 

2. The formula is x = ~ 4 ac . 

2 a 

3. Substituting, g = -(-3)±V(-3)«-4(2)(-5). 


4. 

5. 

6 . .'. 

Check. 


+ 3 ± V9 + 40 
4 _ 

+ 3 ± V49 „ + 3 ± 7 


x = 3 + 7 = 10 nd;c = 3^7 = ^4 = _j 

4 4 2 4 4 

These roots check when substituted in the given equation. 


EXERCISE 142^ 


1. x 2 — 7x + 6= 0 

2 . a: 2 + 2 x - 15 = 0 

3. 2x 2 + 5x+2=0 

4. 3y 2 -7y+2=0 

6. 2 2 2 + 7 2 + 3 = 0 

6. 2 m 2 + 3 w - 2=0 

7. a: 2 — 2 x — 35 = 0 

8. 12 x 2 + 5 x — 2 = 0 

9. 10 m 2 — 11m—6=0 

10. 24 x 2 + 14 x — 3 =0 

11 . x 2 — 2 x — 2 = 0 

12. x 2 + 2 z — 1 =0 

13. 2 x 2 + 3 x — 1 = 0 

14. 3 / - 4 y - 2 = 0 

15. 2 2 2 - 8 2 + 3 = 0 

16. 4 * 2 — 9 < + 2 = 0 

17. 5 c 2 - 11 c - 2 = 0 


18. 

6 m 2 = 5 m 

+ 3 



19. 

1 

TtH 

II 

•b 

1 - 

6 a 



20 . 

8 x 2 + 5 x ■ 

- 1 = 

: 0 


21 . 

2 W 2 + W 

= 1 



22 . 

3c 2 - 2 = 

11 c 



23. 

4 + I = 5l 2 



24. 

0 = 3 y 2 - 

20 + 7 y 

25. 

2x+5 = 

3 x 2 



26. 

x 2 4- -5 x - 

- 1 = 

0 


27. 

1 

kj 

H 

1 

- 5 = 

0 


28. 

y 2 + 3y- 

.4 = 

0 


29. 

2 2 — 5 2 — 

1.5 = 

0 


30. 

w 2 — .3 w ■ 

- 6.8 

= 

0 

31. 

to 

1 

- .3 

= 

0 

32. 

.3x 2 - 5x + .7 

= 

0 

33. 

.8 y 2 - 2.3 y - . 

3 : 


34. 

A w 2 = .3 

- 1.1 

w 











252 


ALGEBRA 


186. # Three methods of solving a quadratic equation have 
been taught: 

a. by factoring, in § 103, page 134. 

b. by completing the square, in § 183, page 246. 

c. by the formula, in § 185, page 250. 


EXERCISE 143 


Solve by the most convenient method. Express surd roots 
correct to tenths. 


1. 

2. 

3. 

4. 

5. 

6 . 

7. 


8 5 


X X 

+ 3 

3 

4 

x + 4 

# — 1 

2 

1 

x — 1 

3 x + 1 

2 

1 


y 4- 4 y - 2 



_L_ + 4 = 
r + 1 r 
s _ s_ _ 
5 - 1 2 “ 


- 2 
1 


8. 

2c _ 

4 j 

c + 5 

c - 3 

9. 

3 m 

2 m i 

2 m + 1 

3 to — 1 

10. 

2 

1 _ 3 

4J/4-1 

5 2y - 1 

11. 

3 z — 1 

2 x + 1 _ 0 


x 4- 1 

£ — 1 



w — 3 
w + 3 


13. 

14. 


7—3 m _ 3 4~ m _ ^ 
2 2 — m 

2c - 1 _ 3c + 1 _ 1 
3 c — 5 4 c 6 


16 . 5i±i + L^2 = 1 

3 < - 3 

16. 3-Z + 5 + 2z + 1 = j 
3 z -f 2 2z 


17. 

18. 

19. 


2a - 3 . a - 3 
a -f 3 2 a + 3 

3r - 1 2 - r 

5 + r 2 r + 1 
5c 3 

2c - 5 c - 4 


20 . 

21 . 

22 . 


6 - 5m 

4- 

co 

m 

2 

2 - 

m 

2s — 5 

s 4' 

1 

3 s 4~ 5 

5s - 

1 

1 + 

1 



-1-+- 1 - 

1 - 2 / 1 — 4 / 


iOIO 






































QUADRATIC EQUATIONS 


253 


EXERCISE 144 

1. If 3 be subtracted from 8 times a certain number, the 
result is 4 times the square of the number. What is the 
number ? 

2. If 7 times the square of a certain number be increased 
by 3, the result equals 10 times the number. What is the 
number ? 

3. There are two consecutive integers whose product is 462. 
What are they ? 

4. There are three consecutive even integers such that the 
square of the first, increased by the product of the other two, 
is 184. What are they 

5. The sum of a certain number and its reciprocal is 
What is the number? (Hint. — Read Note, page 219.) 

6. The denominator of a certain fraction exceeds its numera¬ 
tor by 5. The sum of the fraction and its reciprocal is -Jf. 
What is the fraction ? 

7. The base of a certain triangle exceeds its altitude by 5 
inches. Its area is 88 square inches. What are its dimensions ? 

8. The area of a certain rectangle is 660 square feet. The 
sum of its base and altitude is 52 feet. What are its dimensions ? 

9. The total area of certain two squares is 369 square feet. 
The side of one exceeds the side of the other by 3 feet. How 
long is the side of each ? 

10 . The area of the main waiting room of the Union Railway 
Station at Washington, D. C., is 28,600 square feet. The sum 
of its length and width is 350 feet. What are its dimensions ? 

11. The altitude of a certain right triangle is 5 feet more than 
its base. The hypotenuse is 25 feet. What are its base and 
altitude ? 

12. If a certain number be increased by 3 times its reciprocal 
the sum is \2\. What is the number ? 


254 


ALGEBRA 


187/ Quadratic equations having two unknowns. 

Example 1. Find two integers (whole numbers) such that the 
sum of their squares is 90, and that the larger exceeds twice the 
smaller by 3. 

Solution. 1. Let l = the larger integer 

and s = the smaller. 

2. /. I 2 + s 2 = 90 (1) 

3. and l = 2 s + 3. (2) 

4. Eliminate l by the substitution method of elimination. 

From (2) l = 2 s + 3. (See page 208.) 

Substituting in (1), (2 s + 3) 2 + s 2 = 90. 

5. .*. 4 s 2 + 12 s + 9 + s 2 = 90, or 5 s 2 + 12 s - 81 =0. 

6. Factoring, (5 s + 27) (s — 3) = 0. 

7. s = 3, and s = — 

8. When s = 3, l = 2 ■ 3 + 3, or 9. 

9. — is unsatisfactory as a solution of the problem, since integers 
were desired. 

Check. When s = 3 and l = 9, s 2 + l 2 = 9 + 81, or 90. 

Also 9=23+3. 


When there are two unknown numbers, two independent 
equations must be given. (See page 205.) 

In the following list, one of these equations is a quadratic 
equation and the other is a linear equation. (See page 201.) 


Example 2. 


Solve the system 


\x + y = 2 

\xy = — 15 


a) 

( 2 ) 


Solution. 1. Solve (1) for y in terms of x. y = 2 — x. 

2. Substitute in (2), x (2 — x) = — 15. 

3. :. 2x - x 2 =- 15. 

4. .'. x 2 — 2 x — 15 = 0, or (x + 3) (a; — 5) = 0. 

5. .'. x = — 3, and x = 5. 

6. Since y — 2 — x, when x = — 3, y = 2 — (— 3) = 5; 

when x = 5, y = 2 — 5= — 3. 

Check. Each solution (see page 201) should be checked by sub¬ 
stituting it in both equations, (1) and (2). 

Note. — Not required by either the Regents or the C.E.E.B. 


QUADRATIC EQUATIONS 


255 


EXERCISE 145 


Solve the following systems; check: 


| * “ 2 / = 3 

1 xy = 10 

\x + y= - 2 
1 xy = - 24 
12s + y = 5 
1 xy = 3 
r a 2 + 6 2 = 13 
j a — 2 6= — 4 
f x 2 - y 2 = - 7 
l 2 x — y = — 10 
f cd = - 10 
l 2c + 3d = 11 


f 3 s -f -1 = 0 
1 9 s 2 + 4 t 2 = 5 
j 4 m 2 + n 2 = 10 
i2 m + n =4 
f a: 2 — 2 xy — y 2 — 17 
j x - 2y = 7 
f r 2 + + s 2 = 5J 

1 r — 2 s = 1 


11 . 

12 . 

13. 

14. 

15. 

16. 

17. 

18. 

19. 

20 . 


la 2 + b 2 = 113 
1 a - b = — 1 
j 5 x 2 - 3 y 2 = - 7 
1 y + 2 x = 7 
j m 2 + mw — w 2 = — 19 
1 m — n = — 7 

f * + U = ~ 3 
l xy = - 54 
| z 2 — xy + y 2 = 63 
1 x - y = - 3 
| z 2 + y 2 = 101 
i x + y = - 9 
J z 2 + xy + y 2 = 39 
1 x + y =-2 
j2y + 2x=5xy 
{ 2x + 2y =5 
(3c + 2d=-2 
\ cd + 8 c = 4 
| 7 z 2 + 10 xy = - 8 
| 5 a: + 4 y = — 8 


21 . Find two numbers whose sum is 10 and product 21. 

22. Find two numbers whose sum is 11 and the sum of whose 
squares is 61. 

23. One number exceeds another number by 3. The square 
of the larger exceeds twice the square of the smaller by 14. 
What are the numbers ? 

24. One number exceeds 3 times a smaller number by 1. The 
larger, increased by the square of the smaller, is 11. What 
are the numbers ? 


256 


ALGEBRA 


REVIEW EXERCISE XIII 


1. Divide 2 a 3 + 7 a? — 10 a + 3 by a 2 + 4 a — 3. Check 
by letting a = 2. 

2. Factor : gl 100 x 4 — 49 y 6 c. 4 mz 2 — 9 my 2 

b. (a — 3 b) 2 — 4 x 2 d. y 2 — 16 y + 64 

3. Simplify: ( 4 - + (8 - 

4. Solve and check the system : 

5. Simplify : 6a/108 — 6 aV5J- +V3(5 a — l) 2 . 

6. a. Solve for r : C = — - 

R + r 

b. Find r to the nearest tenth when C = 15.6, R = 4.3. 


6 x — y = .9 
x + 2 y = .8 


7. The numerator and denominator of a certain fraction 
are in the ratio 2:3. If 3 be subtracted from the numerator 
and 6 from the denominator, the value of the fraction becomes f. 
What is the fraction ? 


8. Solve graphically the system : 


2 x — y = 9 

3 x 2 y = — 4 

9. By the formula A =Vs(s — a)(s — b)(s — c) find A 
correct to two decimal places when a = 15, b = 18, and c = 11, 
if s = %{a + b + c). 

10. Find three consecutive even integers such that the 
product of the first and third exceeds 15 times the second by 96. 

11 . What is the formula for the selling price 8 of an article 
which cost M dollars, if the profit is r% of the cost. 

3 a b ,2 ab 


12 . Simplify: 


a -f- b 


+ 


b 2 


Check. 


13. Which of the following numbers are roots of the equation 

2 a; 2 + 5 a; — 3 = 0? (a) 2; (6) - 3; (c) (d) - 1. 

14. How many roots does every quadratic equation have? 
How many roots does every linear equation have ? 










QUADRATIC EQUATIONS 


257 


15. Give the name applied to the 2 in each of the following 
expressions and explain its meaning in each case: 2 x; x 2 ; \/x. 

16. In sending a telegram, there is a fixed rate for the first 
10 words, and a fixed rate for each additional word. If a message 
of 31 words costs 90^, and a message of 45 words costs $1.40, 
what are the two rates ? 

17. The length of a room is 8 feet greater than its width. If 
each dimension is increased by 2 feet, the area will be increased 
by 60 square feet. Find the area of the floor. 

18. A rectangular piece of paper is twice as long as a certain 
square piece of paper, and 3 inches wider. The area of the 
rectangular piece is 108 square inches. Find the dimensions 
of the square piece. 

19. Solve and check: x —- = ---• 

3(x - 1) 2 

20. Find the number whose square equals 20 more than 
eight times the number itself. 

21 . Draw the graph of y = ^ x 2 . 

22. a. Draw the graph of y = mx + b, when m — 2 and 
b = — 5. 

b. Draw the graph also when m = 2 and b = + 5. 

c. What happens to the graph as b increases ? 

23. Given ax + by = 3 

dx + ey = 7 

If x = 2 when y = 3 and x = 5 when y = 8, find a , b, d, and e. 

24. Express in words : 2 n 2 — 3 n + 5. 

25. a. Solve the equation : S = at 2 for t. 

b. Find t, correct to the nearest tenth when S = 250 and 
a = 32. 

26. Is the following statement true or false ? 

A polynomial is an algebraic expression of three terms. 




XV. TRIGONOMETRY OF THE RIGHT TRIANGLE 


188. In applied mathematics, distances and angles often can¬ 
not be measured directly. In many cases the measures can be se¬ 
cured indirectly by measuring other 
distances and angles. 

Angles are measured by means of 
a transit. This instrument contains 
one protractor by which vertical 
angles are measured, and another by 
which horizontal angles are measured. 

The following problem illustrates 
one indirect means of measurement 
of a distance. 

Problem. — 

Find the dis¬ 
tance between 
A and B on 
opposite sides of a stream, without cross¬ 
ing the stream. Fia 2 



Fig. 1 



Solution. 1. Locate C so that 
ZABC = 90°. 

2. Measure CB ; assume it is 60 feet. 

3. Measure ZBCA ; assume it is 50°. 

4. The adjoining scale drawing is then 
made. One space represents 6 feet. 

5. Since AB contains 12 spaces then AB 
represents 72 feet. 

That is, the distance AB is about 72 feet. 

One objection to this method is that 
the scale drawing must be made. Trigo¬ 
nometry is a means of securing this same 
result without use of a scale drawing. 
It is necessary first to learn a few facts about trigonometry. 



Fig. 3 


258 








































































TRIGONOMETRY 


259 


189. The tangent of an angle. 

In the adjoining figure, let Z ABC = 50°. 
represent 1. PiRi is perpendicular to BA. 
BRi form triangle BPxR h in which 
PiRi is the side opposite Z B, and 
BRi is the side adjacent to Z B. 

Observe 

PjRi _ 6 _ t 9 
BRi 5 
and in ABP2R2, 

and in APP3P3, 


Let each space 
P 1 R 1 , PiB, and 


P2R2 

BR2 

P3R3 

BR* 


14 o 

= —, or 1 . 2 ; 


12 

10 ’ 

18 

15 


10 10 

—, or 1 . 2 . 












-T 

z 












f 










c 

Z- L 











Z 










. > 











j 

2 / 

■ - 











/ 











A 











/ 










1 

y 

/ 











A 











/ 











/ 








15 - 

A 


( 

r 



5 


10 




B 





T? 1 



,1 




A: 


42 

Ll 


:3t: 



This fixed value is called 


No matter where P may be on BC, 

the ratio = 1.2, when Z B is 50°. 

BR 

the tangent of Z B and it is generally written tan B 

or tan B — ^ e s ^e opposite Z B in right ABPR 

BR the side adjacent to Z B in right ABPR 


190. When an angle changes, the tangent of the angle changes. 


a. In the adjoining figure, let 
ZCBA = 30° 

Then tan ZCBA = tan 30° 


.57, or .6 


_ AC _ 11.5 
BC 20 

b. Z DBE = 60° 

, DE 17 1 >7 

ten 60 = -g-p = — = i.7 

c. ZFBH = 70° 


tan 70° = - 2 / = 2.75, or 2.8. 

In the table on page 260, the 
tangents of many angles are given, 



correct to four decimal places. 


Thus tan 30° = .5774; tan 70° = 2.7475 

Observe that the tangent increases when the angle increases. 







































































































































260 


ALGEBRA 


Four-Place Table of Values of Trigonometric Ratios 


Angle 

] Sin 

Cos 

Tan 

Angle 

Sin 

Cos 

Tan 

1° 

.0175 

.9998 

.0175 

46° 

.7193 

.6947 

1.0355 

2° 

.0349 

.9994 

.0349 

47° 

.7314 

.6820 

1.0724 

3° 

.0523 

.9986 

.0524 

48° 

.7431 

.6691 

1.1106 

4° 

.0698 

.9976 

.0699 

49° 

.7547 

.6561 

1.1504 

6° 

.0872 

.9962 

.0875 

50° 

.7660 

.6428 

1.1918 

6° 

.1045 

.9945 

.1051 

51° 

.7771 

.6293 

1.2349 

7° 

.1219 

.9925 

.1228 

62° 

.7880 

.6157 

1.2799 

8° 

.1392 

.9903 

.1405 

53° 

.7986 

.6018 

1.3270 

9° 

.1564 

.9877 

.1584 

54° 

.8090 

.5878 

1.3764 

10° 

.1736 

.9848 

.1763 

55° 

.8192 

.5736 

1.4281 

11° 

.1908 

.9816 

.1944 

56° 

.8290 

.5592 

1.4826 

12° 

.2079 

.9781 

.2126 

67° 

.8387 

.5446 

1.5399 

13° 

.2250 

.9744 

.2309 

58° 

.8480 

.5299 

1.6003 

14° 

.2419 

.9703 

.2493 

59° 

.8572 

.5150 

1.6643 

15° 

.2588 

.9659 

.2679 

60° 

.8660 

.5000 

1.7321 

16° 

.2756 

.9613 

.2867 

61° 

.8746 

.4848 

1.8040 

17° 

.2924 

.9563 

.3057 

62° 

.8829 

.4695 

1.8807 

18° 

.3090 

.9511 

.3249 

63° 

.8910 

.4540 

1.9626 

19° 

.3256 

.9455 

.3443 

64° 

.8988 

.4384 

2.0503 

20° 

.3420 

.9397 

.3640 

65° 

.9063 

.4226 

2.1445 

21° 

.3584 

.9336 

.3839 

66 ° 

.9135 

.4067 

2.2460 

22° 

.3746 

.9272 

.4040 

67° 

.9205 

.3907 

2.3559 

23° 

.3907 

.9205 

.4245 

68 ° 

.9272 

.3746 

2.4751 

24° 

.4067 

.9135 

.4452 

69° 

.9336 

.3584 

2.6051 

25° 

.4226 

.9063 

.4663 

70° 

.9397 

.3420 

2.7475 

26° 

.4384 

.8988 

.4877 

71° 

.9455 

.3256 

2.9042 

27° 

.4540 

.8910 

.5095 

72° 

.9511 

.3090 

3.0777 

28° 

.4695 

.8829 

.5317 

73° 

.9563 

.2924 

3.2709 

29° 

.4848 

.8746 

.5543 

74° 

.9613 

.2756 

3.4874 

30° 

.5000 

.8660 

.5774 

75° 

.9659 

.2588 

3.7321 

31° 

.5150 

.8572 

.6009 

76° 

.9703 

.2419 

4.0108 

32° 

.5299 

.8480 

.6249 

77° 

.9744 

.2250 

4.3315 

33° 

.5446 

.8387 

.6494 

78° 

.9781 

.2079 

4.7046 

34° 

.5592 

.8290 

.6745 

79° 

.9816 

.1908 

5.1446 

35° 

.5736 

.8192 

.7002 

80° 

.9848 

.1736 

5.6713 

36° 

.5878 

.8090 

.7265 

81° 

.9877 

.1564 

6.3138 

37° 

.6018 

.7986 

.7536 

82° 

.9903 

.1392 

7.1154 

38° 

.6157 

.7880 

.7813 

83° 

.9925 

.1219 

8.1443 

39° 

.6293 

.7771 

.8098 

84° 

.9945 

.1045 

9.5144 

40° 

.6428 

.7660 

.8391 

85° 

.9962 

.0872 

11.4300 

41° 

.6561 

.7547 

.8693 

86 ° 

.9976 

.0698 

14.3010 

42° 

.6691 

.7431 

.9004 

87° 

.9986 

.0523 

19.0810 

43° 

.6820 

.7314 

.9325 

88 ° 

.9994 

.0349 

28.6360 

44° 

46° 

.6947 

.7071 

.7193 

.7071 

.9657 

1.0000 

89° 

.9998 

.0175 

57.2900 

















TRIGONOMETRY 


261 


191. Using the table of tangents of angles. 

( 

a. To find the tangent of an angle not in the table. 

Example. Find tan 40° 15' 

Solution. 1. tan 40° = .8391'I 

tan 40° 15' = ? > Difference = + .0302. 

tan 41° = .8693 J 

2. 15' is or \ of the difference between 40° and 41°. 

3. Take % of .0302. This is .0075 + 

Add this to .8391, giving .8466, since the tangent increases when the 
angle increases. 

4. /. tan 40° 15' = .8466. 

EXERCISE 141 

Find as above the following : 

1. Tan 20° 4. Tan 25° 30' 7. Tan 75° 12' 

2. Tan 35° 5. Tan 46° 20' 8. Tan 32° 22' 

3. Tan 80° 6. Tan 63° 40' 9. Tan 48° 33' 


b. To find an angle when its tangent is given. 

Example 1. Find Ax if tan x = .6249 

Solution. In the first column headed Tan, .6249 is found opposite 
32°. Therefore Zx = 32°. 


Example 2. Find Ax, if tan x = 1.3560 

Solution. 1. 1.3460 lies between 1.3270 and 1.3764 which appear 
in the second column of tangents. Hence Ax is between 53° and 54°. 
That is: tan 53 <> = x 327 o | Diff. 

tan z = 1.3460 J = .0190 
tan 54° = 1.3764 


} Diff. 
j = .0494 


2. Form the fraction 


.0190 _ 190 

-, or -—— 

.0494 494 


Take of 60' 


m x 60 


This is 23 + minutes J = 23 + 

3. Since the tangent increases when the angle increases then Ax 
must = 53° + 23', or 53° 23'. 



262 


ALGEBRA 


EXERCISE 147 

Find the angle x, when 

1. Tan x = .1405 4. Tan x = .1632 

2. Tan x = .8693 5. Tan x = .4081 


7. Tan x = .7028 

8. Tan x = 1.4608 


3. Tan x = 6.3138 6. Tan x = 1.8295 9. Tan x = .5467 

c. Solving 'problems by use of the tangents of angles. 

Example 3. In the adjoining figure, assume that BC is 


25 ft., and B is 63° 20'. How long is the pole? 

AC 
BC 

tan 63° = 1.9626 
tan 64° = 2.0503 
diff. = .0877 
i X .0877 = .0292 


2 . 


Solution. 1. tan ZB = 

:. tan 63° 20' = 4?, or 
25 

AC = 25 X tan 63° 20' 

3. .*. AC = 25 X 1.9918 

4. .\ AC = 49.79 or about 50' 


/. tan 63° 20' = 1.9626 
.0292 



1.9918 

Note 1 . — ZCBA is called the angle of elevation 
of A at B. 

Note 2. — The value of BC, 25 ft., has two significant figures; for 
this reason, the final result is cut down to two significant figures. 

Example 4. AB represents a lighthouse 250 ft. high. AD 
is an imaginary line parallel to BC. C represents the position of 
a ship. Angle DAC = 31°. Find BC. 

Solution. 1. In such a figure, angle BCA equals angle DAC. 
(Make a tracing of angle DAC on tissue paper, and compare it with 
angle BCA.) angle BCA = 31°. 


2 . 


tan BCA = 


AB 


BC 

3. BC X tan 31° = 250. 

4. .*. BC X .6009 = 250 

5. D. 6 oo9 BC = 416 + ft. 


Tan 31 c 



Note. — Angle DAC is called the angle of depression of point C at 
point A. 









TRIGONOMETRY 


263 


EXERCISE 148 

1. In the adjoining figure, AC is perpen¬ 
dicular to CD ; CD = 100 ft.; angle D = 62°. 

Find AC and AD. 

2. 145 feet from the foot of a high building the angle of 
elevation of the top is 39°. How high is the building ? 

3. From the top of a hill known to be 150 ft. above the level 
of the plain, the angle of depression of a house is 22°. How 
far away is the house from an imaginary point directly below 
the top of the hill ? 

4. At a point 175 ft. from the foot of a building, the angle 
of elevation of the top is 51° 30'. How high is the building? .jj 

5. From a height of 350 ft., the angle of depression of an 
object on the plain below is 28° 15'. Find the distance of the 
object from a point in the plain directly below the point of 
observation. 

6. The angle of elevation of an airplane at a certain point P 
is 38° 30'. Point D, 1500 ft. distant, is directly below the air¬ 
plane. How high is the airplane ? 

7 . At a point 140 feet from the foot of a high chimney, the 
angle of elevation of its top is 40° 30'. How high is the chimney ? 

8. From a hilltop 275 feet above the level of a lake, the 
angle of depression of one sailboat is 42° 20'. The angle of 
depression of a second sailboat directly in line with the first 
boat is 68° 40'. What is the distance between the two boats ? 

9. An observer in an airplane, which is 1500 feet high, finds 
that the angle of depression of a station on the ground is 25°. 
How far distant is he from a point directly over the station ? 

10. When the angle of elevation of the sun is known to be 38°, a 
chimney casts a shadow 125 feet long. How high is the chimney ? 

11. In A ABC, if ZB = 55°, BC = 18 in., and altitude AD 
bisects BC, how long is AH? 




264 


ALGEBRA 


192. Two other ratios of sides of a right triangle are useful 
in solving certain problems. 

193. Sine of an angle. 

a. The meaning of sine of an angle. 

If you measure BP h BP 2 , and 

BPs, carefully, you will find that 

PiRi _ P^Rv _ P ?>Rz 

~BP[ - bp^~~bK 

Wherever you place a point P on 
line BC, when you draw the per¬ 
pendicular PR, the ratio of it to the 
distance BP will always have this same value. This ratio is 
called the sine of angle B (sin B ). In the right triangle BPR, 

sin B = PR = s ^ e °PP 0S ^ e angle B 
BP the hypotenuse of the triangle 

The value of the sine changes as the angle changes. The 

values of the sine for every degree of angle from 1° to 89° in¬ 
clusive are in the columns headed Sin of the table on page 270. 
Thus sin 63° = .8910. 

This means that the side opposite a 63° angle of a right triangle, 
divided by the hypotenuse of the triangle, equals .8910. Again, 
if sin x = .6820, then the angle must be 43°. 

Notice that the sin increases as the angle increases. 

b. Using the table of sines. 

The approximate value of the sine of any angle between 1° 
and 89° can be found by use of this table. 

Example. Find sin 27° 25' 

Solution. 1. sin 27° = .4540 ] 

sin 27° 25' = ? Difference = .0155 

sin 28° = .4695 J 

2- 25' = or ^ of 1°. ^ X .0155 = .0064 

3. Since the sine increases as the angle increases, 

sin 27° 25' = .4540 + .0064, or .4604 



































































TRIGONOMETRY 


265 


EXERCISE 149 

Find : 

1. Sin 48° 4. Sin 35° 20' 7. Sin 20° 5' 

2. Sin 35° 5. Sin 52° 30' 8. Sin 76° 25' 

3. Sin 62° 6. Sin 68° 40' 9. Sin 43° 50' 

10. Find angle x when sin x = .6548 

Solution. 1. .6548 lies between .6428 and .6561 in the column of 
sines. Hence Z.x is between 40° and 41°. 

That is sin 40° = .6428 1 Diff. 

sin z = .6548 J = .0120 
sin 41° = .6561 

2. Write the fraction , or —, and find of 60' 

.0133’ 133’ 133 

4§S X 60 = ^ = 54+ 

3. Since the sine increases as the angle increases then Z.x must be 
40° 54'. 

Find angle x if : 

11. Sin x = .5150 14. Sin x — .3223 17. Sin x = .3500 

12. Sin x = .8988 15. Sin x = .6064 18. Sin x = .8735 

13. Sin x = .9925 16. Sin x = .9624 19. Sin x — .4356 

c. Solving 'problems by means of sines of angles. 



1 Diff. 
j = .0133 


Pole 







266 


ALGEBRA 


194. Cosine of an angle. 

a. The meaning of cosine of an 
angle. 

In the adjoining figure, if you 
measure BRi, BR 2 , and BR 3 , and 
compute the ratios accurately, you 
will find that 

BR\ _ BR 2 _ BR 3 

'bp 1 ~'bF 2 ~JF 3 

Wherever P is taken on BC, the 
ratio of BR to BP will have the 
same value. This ratio is called the cosine of angle B (cos B). 
In the right triangle BPR, BR is the side adjacent to angle B, 
and BP is the hypotenuse. 

cos B = = the side adjacent to angle B 

BP the hypotenuse of the triangle 

The value of the cosine of an angle changes as the angle 
changes. These values are found in the columns headed Cos on 
page 260 for every degree of angle from 1° to 89°. 

Thus, cos 36° = .8090. 

Notice that the cosine decreases as the angle increases. 

b. Using the table of cosines. 

The approximate value of the cosine of any angle between 
1° and 89° can be found by use of this table. 

Example. Find cos 62° 15'. 

Solution. 1. cos 62° = .4695 ] 

cos 62° 15' = ? > Difference = .0155 

cos 63° = .4540 J 

2. 15' is ^ or \ of 1°. £ of .0155 = .0038+. 

3. Since the cosine decreases as the angle increases, 

cos 62° 15' = .4695 - .0038, or .4657. 




































































TRIGONOMETRY 


267 


Find : 

1. Cos 34° 

2. Cos 67° 

3. Cos 46° 


EXERCISE 150 

4. Cos 40° 20' 

6. Cos 50° 30' 
6. Cos 65° 15' 


7. Cos 23° 10' 

8. Cos 72° 5' 

9. Cos 68° 40' 


10. Find angle x if cos x = .6238 

Solution. 1. .6238 lies between .6293 and .6157 in the column of 
cosines, so angle x must lie between 51° and 52°. 

That is cos 51° = .6293 1 Diff. 

cos x — .6238 J = .0055 
cos 52° = .6157 

2. Write the fraction or and find — of 60' 

.0136’ 136’ 136 

^ X 60 = 4£oo, or 24+ 

3. Since the cosine decreases as the angle increases, then angle x 
must be 51° 24'. 

Find angle x if : 

11. Cos x = .9205 14. Cos x = .9588 17. Cos x = .4500 

12. Cos x = .6561 15. Cos x = .4962 18. Cos x = .9310 

13. Cos z = .1736 16. Cos z = .2532 19. Cos x = .7259 


1 Diff. 
j = .0136 


c. Solving problems by means of the cosines of angles. 


Problem. A 22 ft. ladder leans against a house. Its foot is 
12 ft. from the house. What angle does it make with the 
ground ? 


Solution. 1. cos B = = .5454 

2. cos 56 = .5592 1 1 

cos B = .5454 / • d [ .0146. 

cos 57 = .5446 J 

3 . i 3 | x 60 ' = 56 

4 . .*. B = 56° 56', or about 57°. 











268 


ALGEBRA 


EXERCISE 151 


(You will need to use a sine or cosine in each example.) 


1 . A boy is flying a kite at the end of 500 ft. of twine. 
Directly below the kite a second boy stands. The angle made 
by the twine (assumed to be stretched straight) with an imagi¬ 
nary line running from the first boy to the second is approxi¬ 
mately 50°. Determine the approximate height of the kite. 

2 . In triangle ABC, BC is 18 ft., angle B is 70°, and AB is 
8 ft. Draw a triangle to represent these 
conditions and draw the altitude AD. 

a. Find the length of AD. (Use the 
sine.) 

b. Having found AD, find the area. 

3. In the figure at the right is a 
right triangle in which AB = 10 ft., and 
BC = 15'. 



а. Find angle C. b. Find AC. 

4. In triangle ABC, in which angle 

C is a right angle, and angle B is an 
angle of 24° 45', side AB is 24 in. long. 
Find the length of AC and of BC. 

б . In the adjoining figure, OC is 
perpendicular to AB. 

а. Find AC. b. Find OC. 

б . In the adjoining figure, ABCD 
is a rectangle. 

a. Find ZCDB. 

b. Using ZCDB, as found, find 
DB. 



7. In the adjoining figure, AD is 
perpendicular to BC. 

a. Find AD. 

b. Find the area of A A BC. 



D 27 








TRIGONOMETRY 


269 


EXERCISE 152 

Miscellaneous Trigonometric Exercises 
Find the missing numbers in the following examples : 


Ex. No. 


In Triangle ABC , in 

which Angle C = 

= 90° 

BC 

AB 

AC 

Angle B 

Angle A 

1. 

846 

? 

625 

? 

? 

2 . 

975 

1154 

? 

? 

? 

3. 

? 

683 

527 

? 

? 

4. 

? 

? 

962 

36 ° 30 ' 

? 

5. 

? 

378.5 

? 

42 ° 

? 

6 . 

265.8 

? 

425.3 

? 

? 

7. 

483.7 

1292 

? 

? 

? 

8 . 

37.25 

68.43 

? 

? 

? 

9. 

? 

29.75 

18.54 

? 

? 

10. 

? 

361.7 

284.3 

? 

? 

11. 

75 

? 

60 

? 

? 

12. 

85 

100 

? 

? 

? 

13. 

? 

90 

60 

? 

? 

14. 

? 

? 

120 

62 ° 

? 

15. 

? 

150 

? 

36 ° 15 ' 

? 


16. At a time when the angle of elevation of the sun is known 
to be 25° 15', a chimney casts a shadow 85 feet long. How 
high is the chimney ? 

17. In rt. A ABC, having Zi = 90°, ZB — 28°, and 
AB = 25 in., how long are : AC, BC ; the altitude AD to 
BC; and RD? 

18. At a point 145 ft. from the foot of a tower, on which 
stands a flagpole, the angle of elevation of the top of the tower is 
35°, and of the top of the pole 47°. How high is the tower, 
and how long is the pole ? 














270 


ALGEBRA 


19. A ladder 16 ft. long leans against a building. If it makes 
an angle of 58° with the ground, how high on the building does 
it reach ? 

20. How long must a guy wire be to reach from the top of a 
25 ft. telephone pole to a point on the ground 20 ft. from the 
foot of the pole, and what angle will it make with the ground ? 

21. From a window 35 ft. above ground level, the angle of 
depression of an object on the ground is 40°. How far is the 
object from a point on the ground immediately below the 
observer ? 

22. Charles, on one side of a stream, holds a 10 ft. rod in 
vertical position, with one end on the ground. Edward, on 
the other side, measures and finds the angle of elevation of the 
top of the rod is 15°. How far apart are the two boys? 

23. A hill has a slope of about 15°. From the foot of the hill 
to its top is 350 ft. How high is the hill ? 

24. From 500 ft. in the air the angle of depression of a point 
on the ground is 38°. How far is the point from directly below 
the observer ? 

25. In triangle ABC, AB is 18 in., Z B is 45°, and BC is 25 in. 
Find the altitude from A to BC, and the area of the triangle. 

26. In rectangle ABCD, the side A B is 12 in. and BC is 20 in. 

a. How long is diagonal AC? 

b. How large is the angle made by AC and BC ? 

c. How large is the angle made by AC and AB? 

27. An observer measured the angle of elevation of a hill at 
one point and found it to be 60°. Then he walked back 100 ft. 
from the hill in a straight line running through his first point of 
observation, measured the angle of elevation and found it to be 
40°. How high was the hill ? 

Suggestion. — 1. Represent the height of the hill by x ft. 2. Represent 
the distance of the first point of observation from immediately below the 
hilltop by y ft. 3. Make one equation, involving x and y , using the angle, 
60°. 4. Make a second equation, involving x and y, using the angle, 40°. 


TRIGONOMETRY 


271 


28. At 120 ft. from the foot of a tower, the angle of elevation 
of the top was 64° 30'. How high was the tower if the tele¬ 
scope of the transit was 5 ft. above ground level ? 

29. If the angle of elevation of the sun is 70° 15', what is the 
height of a telephone pole which casts a shadow 20 ft. long ? 

30. In a figure like that for example 5, page 278, how long is 
OA, if AC is 4 in. and angle AOC is 22^-° ? 

31. Repeat example 30, if angle AOC is 18°. 

32. Let y = sin x. Draw the graph of this relation. Use for x, 
the values 1°, 10°, 20°, 30°, 40°, 50°, 60°, 70°, 80°, and 89°; place 
them on the horizontal axis. Take the values of the Sin from 
the table on page 270, writing them correct to two decimal places. 

33. From the top of a hill which is 55 ft. above lake level, the 
angle of depression of a rowboat on the lake is 18° 30'. How 
far (from directly below the observer) is the boat ? 

34. How long must a ladder be to reach from a point on the 
ground 15 ft. from the side of the building to a point 23 ft. up 
on the side of the building ? 

35. The height of a tower was observed at two points which 
were on the same level with and in the same straight line with 
the foot of the tower. At the nearer point, the angle of elevation 
of the top of the tower was 48° 30' and at the other it was 37° 45'. 
If the two points were 50 ft. apart, how high was the tower ? 

36. In a circle whose radius is 15 in., chord AB makes an 
angle of 40° with the radius drawn to B. How long is the 
perpendicular from the center of the circle to the chord, and 
how long is the chord if this perpendicular is known to divide 
the chord into two equal parts ? 

37. In A ABC, ZB measures 110°; AB is 12 in. long and 
BC is 20 in. long. How long is the altitude drawn from A to 
side BC, and what is the area of A ABC? 

38. Draw a A ABC having Z B an acute angle. Represent 
the length of AB by c and of BC by a. Express the length of 
the altitude from A to BC by a formula in terms of a, c, and Z B. 


XVI. VARIATION 


195. The purpose of this chapter is best made clear by an 
example. 

You know that the area of the rectangle of altitude 5 and 
base b is given by the formula A = 5 b. You know that b can 
be any number; that there is a definite value of A for every 
value of b ; that A changes as b changes ; that there is a definite 
value of b for every value of A ; that b changes as A changes. 
The fact is that A and b change together, under the control of 
the relation A = 5b. 

A and b are variables, since each may have many different 
values in the formula. They are called related variables, since 
they vary together. 5 is called a constant since it is 5 and only 
5 throughout the problems covered by the formula. 

If a workman receives a fixed wage per day, the total wages due 
him changes from day to day if he works and remains unpaid. His 
daily wage is constant; his total wage is variable. 

Scientists often are not interested in the definite values of the 
numbers in such a formula. Instead they want to know what 
is the effect upon one of the numbers of a change of some sort 
in the other. 

Thus, in the formula S = % gt 2 , g and 2 are constants ; S and t are 
related variables, for obviously S changes when t changes. A scientist 
wants to know, how does S change when t changes? Does it change 
more or less rapidly? etc. 

This chapter is concerned with the changes or variations in 
two or more numbers which change together according to some 
mathematical law. 


272 


VARIATION 


273 


196. The graph of the formula A = 5 6 . 

A 


When b = 

0 

2 

4 

6 

8 

10 

then A — 

0 

10 

20 

30 

40 

50 


a. It is clear that: 

A increases when b increases 
A decreases when b decreases 

b. When b = 2 , A = 10. 

When b = 4, A = 20. 

Observe that A is doubled if b is 
doubled. Similarly, if b is trebled, 
then also A is trebled. 

c. Observe that J -£ = 3 £- = - 3 /, etc. 

That is, the ratio of any value of A to the corresponding 

value of b is always 5. This fact can be inferred directly from 



123456789 10 


the formula. 


For, since A = 5 b, then — = 5. 

b 


We say: A varies directly as b. 

197. While any formula can be studied in this manner, all 
these facts are true for any numbers x and y satisfying an equa¬ 
tion of the form x = my, or - = m. 

y 

One variable varies directly as another if the quotient of any 
value of the one divided by the corresponding value of the other 


is constant, as x and y above. / 

Then such facts as the following are true: 

a. x increases when y increases; 

b. x decreases when y decreases; 

c. x is doubled when y is doubled; or 

d. x is halved when y is halved; etc. 

Example. The perimeter, p, of a square of side s is given by the 
formula p = 4 s, from which 2 = 4. Therefore, p varies directly as s. 


























































































274 


ALGEBRA 


198. A slightly different type of variation. 

Example. a. The formula for the volume of the rectan¬ 
gular parallelepiped whose base is a square of side s and whose 
altitude is 3 is V = 3 s 2 . 

b. Make a table of corresponding values of s and V and draw 
the graph representing V = 3 s 2 . 



When s = 

0 

2 

4 

6 

8 

10 

Then V = 

0 

12 

48 

108 

192 

300 


c. Observations. 


1 . 

2 . 


3. 


V increases when s in¬ 
creases, but more rapidly 
than s. 

When s = 2, V = 12. 
When s = 4, V = 48. 
When s is doubled, V is 
multiplied by 4. 

12_12_q.48 = 48 . 

2 2 4 ’ 4 2 16 


108 

6 2 


108 = 
36 


etc. 


Therefore V is proportional 

to s 2 . This fact can be inferred directly from the formula. For, since 


V = 3 s 2 , then - = 3. 

s 2 


We say, V varies directly as the square of s. 


In general, one number varies directly as the square of another 
if the quotient of any value of the one divided by the square 
of the corresponding value of the other is constant. 


Example 1. Since the area, A, of a square of side, s, is given by the 

formula A = s 2 , then — = 1. 

s 2 

Therefore A varies directly as the square of s. 

This means that A is multiplied by 4 when s is doubled. 

that A is multiplied by 9 when s is trebled; etc. 






































































































VARIATION 


275 


199. Another type of variation is illustrated by the formula 
I = PRT. Let T = 2. Then 1 = 2 PR. 

Since 7=2 PR, then = 2 . Hence, in general I is pro- 
PR 

portional to the product of P and R. 

We say, I varies jointly as P and R. 

In general, one number varies jointly as two or more others 
when any value of the one divided by the product of the cor¬ 
responding values of the others is constant. 

Example. Since V = Iwh, = 1 . Hence V varies jointly as 
Iwh 

l, w, and h. 


EXERCISE 153 


Complete the following sentences : 

A 

1. Since A = ab, and therefore —- = ?, then A varies 

ab 

-as a and b. Hence, if a alone is doubled, then A 

is-; if b alone is trebled, then A is_; 

if a is doubled and b is trebled, then A is_ 


2 . Since A = ab, if a = 2 , then 


A 

b 


? Hence A varies 


.as_If b is halved, then A is 

If A is doubled, then b is_ 


A 

3. Since A = 7 rr 2 , then — = ? Therefore A varies as 

r 

__Hence, if r is doubled, A is multiplied by 

_; if r is trebled, A is multiplied by_ 

4. Since S = 4 rr 2 , then S varies_as_ 

Hence if r increases, S _, only_rapidly; 

if r is doubled, then S is_ 


5. Since V = f tv 3 , then S varies_as_ 

Hence if r increases, V _, only___rapidly; 

if r is doubled, V is multiplied by_ 























276 


ALGEBRA 


200. A quite different type of variation is illustrated by the 
following example. 

Example. a. In the formula A = hb, for the area of a 
rectangle, let A = 20. Then hb = 20 is the formula for all 
rectangles having an area 20. 

b. Make a table of corresponding values of h and b and draw 
the graph of the relation hb = 20. 


When b = 

1 

2 

4 

5 

10 

2 

20 

Then h = 

20 

10 

5 

4 

1 


c. Observations. 

1. When b increases, h de¬ 
creases. 

2 . When h increases, b de¬ 
creases. 

3. When b = 2, h = 10. 

When b = 20, h = 1. 

So, if b is multiplied by 10, h 
is divided by 10. Similarly, if 
b is multiplied by 5, h is divided 
by 5. 

We say h is inversely proportional to b, and that b is inversely 
proportional to h. 

In general, one number varies inversely as another when the 
product of any value of the one and the corresponding value of 
the other is constant. 

Example, p = br is a well-known formula. Suppose we 
consider all cases in which p is constant, — like 100, or 50, or 
80, etc. Then br = a constant. 

Hence b varies inversely as r; that is, the base increases if 
the rate decreases, and vice versa; if the base is doubled, the 
rate is halved; etc. 









































































VARIATION 


277 


EXERCISE 154 

Miscellaneous Examples of Variation 

1 . If V = Iwh and l is constant, then V varies_ 

as w and h. If then w alone is doubled, V is_; if 

w is doubled and h is trebled, then V is_ 

2 . If V = Iwh and l and h are constant, then_ 

3. If I = PRT, where 7, P, R, and T a fe variable, then 7 

varies_as_ 

4. If 7 = PRT, and if R and T are constant, then I varies 

-as-In this case if I is multiplied 

by 5, then_is_by_ 

5. If 7 = PR T, and 7 and T are constant, then_ 

varies-as_In this case, if P is 

increased, then_is_ 

6 . S = % CL is a formula in geometry. If S, C, and L are 

variable, then S varies_as_If L is 

constant, then S varies_as_If 

S is constant, then C varies_as_, so 

that when C increases, then__ 

7. C = nc is the formula for the cost of n articles at cost c 

per article. If n is constant, the total cost varies_:_ 

as_ (statejt in w o rds) _If c is constant, 

then the total cost varies_as_ (state it m words) _ 

If C is constant, the number of articles (n) varies_ 

as_ (state it in words) _; so that the number of 

articles decreases as the cost per article_ 

8 . The intensity of light varies inversely as the square of the 
distance from the source of light. Select appropriate letters to 
represent these variables, and express this relation by a formula. 

9 . The per capita cost of instruction of pupils in a room 
varies inversely as the number of pupils in the room. Select letters 
to represent the variables, and express this relation by a formula. 































SUPPLEMENTARY EXERCISES 


EXERCISE 155 


1 . By the formula A = 1C, find l when A= 422.1, and 
C = 23.45. 

2 . By the formula S = 2 x rh, find r when S = 1188,x = 2 ^-, 
and h = 21 . 

3. By the formula A — \ a(b + c), find a when A = 135, 
b = 15, and c = 21. 

4. By the formula S = J gt 2 , find <7 if 8 = 257.6 and t = 4. 
(Remember that t 2 = t • t.) 

5. By the formula ^4 = J ab, find a when A = 403 and 
b = 31. 


6 . By the formula V = Iwh, find h when V = 3960, l = - 3 ^-, 
and w = 5. 

7 . By the formula F = irr 2 h, find h when V = 264, 
x = and r = 3. 

8 . What is the area of the circle whose diameter is 15 in. ? 


9. What is the simple interest on $1650 at 7% for 2 yr. 4 mo. ? 

10 . What is the interest on $2500 at 6 % for 3 yr. 9 mo. ? 

11 . F = W/r is a formula used in physics. 

a. Find F when W = 250 and r = 5. 

b. Find W when F = 700 and r = 6 . 


12 . K = J Mv 2 is a formula used in physics. 
Find K when M = 200 and v = 25. 

13 . H = .24 c 2 Rt is a formula used in physics. 
Find H when t = 120, R = 12 , and c = 10. 


E 


is a formula used in the study of electricity. 


14 . C = 

R + r 

Find C when E = 1.08, R = 220, and r = 10. 


278 



SUPPLEMENTARY EXERCISES 


279 


EXERCISE 156 

1. By the formula F = Iwh, find h when F = 672, l = 12 , 
and w = 8 . 

2 . By the formula V = Iwh , find w when F = 2340, / = 20, 
and h = 13. 

3. By the formula V = Iwh , find l when F = 1360, w = 8 . 5 , 
and ^ = 10 . 

4. By the formula A = Ic, find c when ^4 = 3309.6 and 
l = 12 . 

6 . By the formula 8=2 7 rr^, find h when S = 528, 7 r = 
and r = 14. 

6 By the formula A = \hb, find b when A = 189 and 
h = 14. 

7. By the formula V = %hb, find b when F = 247 and 
h = 19. 

8 . By the formula F = % hb, find h when F = 232 and 
b = 29. 

9 . By the formula / = PRT, find T when I = $273, 
P = $650, and R = 7 %. 

10 . By the formula I = PRT , find P when I = $555, R = 6 %, 
and T = 2^ yr. 

11 . What is the altitude of the rectangle whose area is 594 
sq. ft. and whose base is 33 ft. ? 

Solution. — 1. The formula for the area of a rectangle is A = ab. 

2. In this problem, A = 594; b = 33; a — ? 

Complete the solution as in Examples 1-10. 

12 . What is the base of the triangle whose area is 651 sq. ft. 
and whose altitude is 21 ft. ? 

13. What is the altitude of the parallelogram whose area is 
153 sq. yd. and whose base is 8.5 yd. ? 

14. What is the altitude of the rectangular solid whose 
volume is 3240 cu. in., whose width is 15 in., and whose length 
is 24 in. ? 


280 


ALGEBRA 


201. Addition and subtraction of terms having literal coeffi¬ 
cients. 

The terms ax and bx are like with respect to x, since they 
have the common factor x. The coefficients of x in these terms 
are a and b respectively. 

Just as 3 x + 4 x = (3 + 4) x, or 7 x, so ax + bx = (a + b) x. 

Just as 7 x 2 — 9 x 2 = (7 — 9) x 2 , or — 2 x 2 , so ax 2 — bx 2 = (a — b) x 2 . 

These are examples of the rule: to add like terms , multiply 
their common factor by the algebraic sum of its coefficients. 


EXERCISE 157 

Find the sum: 


1 . 

2 . 

3. 

4. 

6 . 

6 . 

ax 2 

mxy 

2 ctf 

ax 3 

Atf 

ax n 

bx 2 

— nxy 

% 

CO 

— bx s 

- Btf 

g l 

iO 

7. 

8 . 

9. 

10 . 

11 . 

12 . 

ax 

mx 2 

rxy 

2 ay* 

Ax 3 

cx m 

— bx 

- nx 2 

sxy 

1 

CO 

O' 1 

CO 

- Bx 3 

— dx m 

cx 

1 

- txy 

4 cy 2 

Cx 3 

— ex m 


13 - 18 . In Examples 1-6, subtract the lower number from 
the upper. 

19. Add ax + by + cz and dx — ey +/z. 

20 . Add 2 ax + 3 by + 4 cz and dx — 5 ey + 3 fz. 

21 . Add ax 2 + bxy + cy 2 and Ax 2 — Bxy + Cy 2 . 

22 - 24 . In Examples 19-21, subtract the second polynomial 
from the first. 

25. Subtract ax + by — cz from bx + ay — cz. 

26 . Subtract ax — by — cz from Ax + By + Cz. 

27 . From mx 2 — nx + p subtract rx 2 — nx — p. 

28 . From ax -f by + c subtract Ax + By — C. 

29 . Subtract 5 Ax + 4 By + 6 Cz from 2 ax + 3 by —4 cz. 

30 . From ax 2n + bx n + c subtract 2 x 2n — 3 x n + 4. 












SUPPLEMENTARY EXERCISES 


281 


EXERCISE 158 

Remove parentheses and combine terms: 

1. 6 c - 58 c - (10 c - 5) - 9| 

2. (8 t - m) - [9 t - (5 t - 3 m) - 4 1] 

3. (4x -2y) - (fix - [5a + 7y -2] -Sy) 

4. 12 c - J6 c — [9 c + 4] + [5 — 6 c]J 

5. 2 n — [Sn — J4 w + 4J — J — bn — 9j] 

6. a — (2 a — [3 a + 2] — [4 a — 5]) 

7. c — [2 c — (6 a — b) — (c — 5 a — 2 6)] 

8. 3 n — {4 n — (2 w + [5 n + 6]) j 

9. 5 r — j6 r — (7 r — [3 r — 2])} 

10. (8 t - 4) - [(5 t + 1) + (3 t - \2 t - 1{)] 

11. (11 2 - 6) - [(3 z - 2) - (4 z + {6 z - 8})] 

12. (2 r - 5) - l - (3 r + 2) - (7 r - [8 r - 1])} 

13. m — [n + (3 m — 4 n)] — [2 m — 3 n — (m + 3 n)] 

14. (2 a; - 3 y) — (4 y — [5 x — ?/] + [6 x — 4 ?/]) 

15. 3 - {(5 ab - 8) - (4 ab - 16) + (2 ab - 7)} 

EXERCISE 159 

Find the sum of: 

1. ax, bx, and cx 3. 2 ax 2 , — 3 bx 2 , ex l , — 4 dx? 

2. mx, nx, and — hx 4. 5 mx, —2 nx, —6px, + 8rx 

6. 2(x + y), - 5(x + y), and S(x + y) 

6. — 3 (m + n) 2 , + 4 (m + n) 2 , — 5(m + n) 2 

7 . Add S(x - y) - 4(a; - 2 ) + 5(y - 2 ), - 2(a; - y) + 

6 (z — 2 ) — 3(i/ — 2 ), and (x — y) — 2(x — 2 ) + 4(y — 2 ) 

8. Add 6(x + y) 2 - 5(x + y) - 4, - 4(* + 2/) 2 + 2(a: + 2/) 
H- 3, and + 3(:r + y) 2 — {x + y) — 2. 

9. Add 8 (r + 5 ) + 5 (t + s) - 2(r + <), “ 7(r + s) + 

5 (t + s) + 3(r + <), and 2(r + s) — 9 it + s) - 5(r + /). 

10. Add ax + by + C 2 and + c?/ — fz. 


282 


ALGEBRA 


202. Multiplication of terms having literal exponents, or 
decimal coefficients. 

Just as x 7 • x 9 = x 7+9 , or x 16 , so x a - x b = x? + 6 . 

Similarly x m {x n — x*> + x h ) = x m + n — x m + » +6 . 

Also x 7a — 2/ 30 

X 3a 4 . yla 

x? a — x 3a y 3a This is x Za (x‘ 2a — y 30 ) 

_ + x 2a y 2a — y^ This is y 7a (x 2a — y 3o ) 

gtffci _ x^y^ 1 "l - x^ a y^ a _ 2 /^° 


EXERCISE 160 


Find the following indicated products : 


1. 

a m • a 4 5. x° • 

a: 6 


2. 

x c • x 5 6. y 2a 

V 


3. 

y n • y n 7. z 2a 

■ z 30 


4. 

*1 

N 

8? 

00 

'I 

pb 


13. 

(3 x p y q )(+ 4 x p y q ) 


29. 

14. 

(14 x n y m ) (— 3 x 2n y Sm ) 


30. 

15. 

(13 a 2 6 3 ) (— 2 a m b n c n ) 


31. 

16. 

(x 2n+1 y) (x 2 y n ~ l ) 


32. 

17. 

x 2 (x n — x m + x p ) 


33. 

18. 

(— £ a ) (a: a — x b + z c ) 


34. 

19. 

2 / n fe m - y p + 2) 


35. 

20. 

a 2 6 3 (a n — 6 m ) 


36. 

21. 

(— 2 a fc 6) (3 a 2 — 4 6 3 ) 


37. 

22. 

(-5 t n r m )(t n - b m ) 


38. 

23. 

x n (x n — x n ~ l + x n ~ 2 ) 


39. 

24. 

- y\y m + y m ~ l + y m ~ 

->) 

40. 

25. 

— x n (x n — x 2n + X Zn ) 


41. 

26. 

2 r z {r m - 3 r n + 4 r p ) 


42. 

27. 

ir?/(ic n — 7/ m ) 


43. 

CO 

Cl 

ab(a 2m - 2 a m 6 n + 6 2n 

) 

44. 


9. x m y n • x m y n 

10. a n b m • a 2 b 3 

11. (-2a")(+3a») 

12. (-4^)(+5a:«) 
(a m — b m )(a m — b m ) 

(a* + b m )(a n — b m ) 

(x n + 5)(x n — 10) 

(; y m - 6)(y m + 6) 

(a* + 2 b m )(a n - 3 b m ) 
(D» - 7)(Z)» - 3) 

(r 2n _ 6 r » - 3)(r“ + 3) 
(a 2 * -3 a* - 5) (a* - 4) 
(y 2c + 5 y e — 7)(y e - 3) 
{k il - 8 k 2t + 4)(i k 2 ‘ - 4) 
(V 2r — 6 V+ U)(V- 2) 
(x >n — x ln — x n ')(x n — 3) 
(- .7 a;*)(+ 3 a;*) 

(- 1-5 i/)(+ -5 y 2 ) 

(.5 x 2 — .25 x+l)(.5 a;+l) 
(.6 x 3 - .3 x 1 + .2)(.3 a;”) 




SUPPLEMENTARY EXERCISES 


283 


EXERCISE 161 * (Chiefly Optional Factoring) 


Find all the prime factors^f the following. (Review § 95.) 
1. 3 s 3 - 13 s 2 -10 s * 30. 42 t - 15 tx - 3 tx 2 

31. 6 r 4 + 7 r 3 - 5 r 2 


2. 18 at 2 + 48 at -|- 32 a 

3. 20 m 2 r - 125 n 2 r 

4. 18 a 2 c 2 - 57 ac 2 +45 c 2 

5. 9 A 3 - 16 n 3 

6. ax 2 — 2 ax —35 a 

7. 3 c 2 + 4 c - 7 

8. 6 s 2 t — 7 st — 5 t 

9. 16 a 4 - 80 a 2 + 100 

10. 3 x 2 + 9 x - 120 

11. 5 r 2 + 16 r + 3 

12. 18 x 2 y 2 - 3 xy 2 - 10 y 2 

13. 3 m 2 — 30 m + 75 

14. 2 n 2 — 14 n — 36 

15. 18 x 2 + 17 x — 15 

16. 15 t 2 — 6 t — 21 

17. 4 rV + 4 rs 2 — 15 s 2 

18. a 4 — c 2 

19. 5 pr 2 + 5 pr — 10 p 

20. 3 z 2 + 7 + 22 z 

21. 24 m 2 a + 18 ma — 15 a 

22. 18 6c 2 — 2 for 2 

23. 3 x 4 — 17 x 3 - 56 x 2 

24. 45 r 4 ^ — 80 r 2 s 5 

25. 169 a 2 + 78 a& + 9 b 2 

26. 3 fc 2 + 33 k + 72 

27. 9 x 2 — 4 xy — 13 y 2 

28. (P‘6? -\- 0 cd — 52 

29. z 2 + | x + A 


32. 6 ma: 2 — 13 mxy + 6 my 2 

33. 6 a: 2 + 23 a: 3 - 18 x 4 

34. 4 x 2 ?/ — 12 xy 2 + 9 y 3 

35. 21 a 2 - 10 b 2 - 29 ab 

36. m?x 2 — 7 mx — 44 

37. 2 am 2 — 3 aw —20 a 

38. 169 m 2 — 26 mn + n 2 

39. w 6 + 72 - 18 uP 

40. 8 a 4 + 18 a 2 - 35 

41. 4 (Pd — 8 cd — 21 d 

42. 4 a 2 y 2 — 28 ay 2 + 49 y 2 

43. z 2 + 2 zt — 63 ^ 2 

44. 20 c 2 - 30 cd - 140 d 2 

45. 10 x 2 — 13 xy — 30 y 2 

46. 9 a — am 2 

47. a 2 — 36 b 2 + 5 

48. 9 x 2 + 3 x — 56 

49. 10 a 2 + 19 a — 15 

50. 12 x 4 - 8 x 3 - 4 x 2 

51. 15 (Pd — 10 Pd — 25 cd 

52. 4 x 4 y 2 + 2 x s y s — 7 x 3 t/ 2 

53. 55 m 2 x 5 + 2 mx 5 — 21 x 5 

54. b — 16 + 15 b 2 

55. - 25 x + x 2 + 100 

56. 18 - 19 c + c 2 

57. - 50 a + a 2 + 49 
68. 196 y 2 - A 5 ¥ a 4 


284 


ALGEBRA 


EXERCISE 162 * 


Optional Factoring Involving Literal Exponents 


Example 1. £ 2 ® — 18 + 81 

Example 2. 64 x in — 121 y 6m 

Example 3. 3 c 2fc + 4 c k — 7 

Factor and check: 

1. a+ — 14 x 2k + 49 

2. m 6 a _ 24 m 3o _|_ 144 

3. 25 y 2n — 169 w* m 

4. 5 y 2k — 10 y 2k z 

5. 7 a 2x — 14 a 2x 6 + 49 a 2x b 2 

6. 7 k 2a - 14 k 2a R + 7 k 2a R 2 

7. .4i/ TO - 1.2 y 2m + 1.6 

8 . 5 yZm _ 25 _ 37 ylm 

9. z 2a + 2z a 6 a + fc 2a - 16s 2a 

10. 36 x 2m - 9 ?/ m 

11 . 2 - y c - y 2c 

12. 2 + - 6 x 2n 

13. a 2 — 2 ab + b 2 — ?/ 4c 

14. x 2a - 8 + 12 

15. x* n + 17 x 2n - 60 

16. 400 R 2x - 169 T 2y 

17. 169 - 26 R 2n + R 4n 

18. a 8m + 6 a* m b - 91 b 2 

19. x 2m + 11 x m - 12 

20. a 2n + 18 a n + 65 

21. 2 n 2x -14 n x - 36 

22. 5 pr 2y + 5 pr v — 10 p 

23. y 2k -16 y k + 55 

24. 66 — 5 x n + x 2n 


■ ( x a - 9) {x? — 9). 

(8 x 2n — 11 y 3m )(S x 2n + 11 y zm )• 

(3 c k + 7)(c* — 1). 

25. (x + y) 2k — (a — b) 2k 

26. (x— y) 2k —2(x— y) k w k +w 2k 

27. m 2n — 4 m n + 4 — 9 

28. 5 2 * - 8 2r 

29. a ix -2a 21 - 48 

30. 100 x 4n - 49 y« m 

31. 12 c 4p + 7 c 2 ? - 12 

32. 3 y 2a - 6 y a 

33. 1/ 20 -50 y a + 625 

34. a 3 6 2p - 81 a 3 

35. 5 y 2n + 10 y n - 120 

36. 4 x 2n — 15 x n i/ n — 4 ?/ 2n 

37. R 2 - T 2 + 6 TAP- 9 M 2x 

38. 9 a: 2 ° + 4 2 /V - 4 z 2c - y 2b 

39. ax 3 — 6 ax 2 y b + 9 axy 2b 

40. 3 x 2m — 2 x m y n — 8 t/ 2n 

41. 4 a 2x + 5 2 - 4 a x b — y 2c 

42. w 6x + 72 — 18 w 3x 

43. 4 c 2x d — 8 c x d — 21 d 

44. 4 a 2p 2/ 2 — 28 a v y 2 + 49 y 2 

45. (x + y) 2c — 25 

46. (m — n) 2x — 144 y 2a 

47. C 2x + 2 C X D X + D 2x - 49 

48. x 2a — 8 x a y + 16 2 / 2 — 9 z 2e 


SUPPLEMENTARY EXERCISES 


285 


EXERCISE 163 

Miscellaneous Factoring Involving Decimals 
Example 1. x 2 — .09 = {x — .3)(x + .3). 

Example 2. x 2 — x + .25 = (x — .5)(x — .5). 

Example 3. .3 x 2 — 1.5 x + 1.8 = .3(x 2 — 5 x + 6) 

= .3(x - 3)(z - 2). 

Factor and check by multiplying the factors : 


1 . x 2 - .36 

2 . y 2 — 1.4 y + .49 

3. m 2 + 1.6 m + .64 

4. .11 x 2 - .22 x + .11 

5. .25 a 2 - .81 b 2 

6 . .25 + y + y 2 

7. .01 c 2 - .0025 d 2 

8 . y 2 - .8 y + .12 

9. c 2 - 1.2 c + .36 

10 . m 2 — 6.25 n 2 

11 . k 2 - .7 k + .12 

12 . .3 y 2 - 1.2 y + 1.2 

13. .25 y 2 - .25 z 2 

14. b 2 — .7 b + .1 

15. .09 B 2 - .64 C 2 

16. 7.4 M 2 - 7.4 iV 4 

17. iA 2 - ^Y 2 

18. y 3 - .16 y 2 

19. 121 C 2 - 1.44 D 2 

20 . .2 n 2 — 1.4 n — 3.6 

21 . .3 m 2 — 3 m + 7.5 

22 . 1.5 t 2 - .6 t - 2.1 

23. .5 x 2 — x — 17.5 

24. .7 x h — .7 x 


25. c 2 - .1 c - .06 

26. .04 a 2 + .12 ab + .09 b 2 

27. .36 m 2 + .60 mn + .25 n 2 

28. .16 ,4 2 - .56 + + .49 

29. 1.21 C 2 — 1.44 M 2 

30. .5 x 2 — 3 x + 4.5 

31. .6 ,4 4 - 9.6 £ 4 

32. M 2 - .2 M - .15 

33. x 2 — x + .16 

34. k 2 -1.2 k + .35 

35. .0004 y 2 - .0009 

36. .5 hx + .5 hz — .5 h 

37. .25 x 2 - .49 y 2 

38. y 2 - .7y + .1 

39. R 2 — AR — .21 

40. B 2 + .8 B - .09 

41. 6.25 x 4 - .81 y 4 

42. .16 a 4 - .8 a 2 + 1 

43. .6 s 2 t — .7 st — .5 t 

44. .03 x 2 + .09 x - 1.2 

45. 1.69 m 2 — .26 mn + .01 n 2 

46. .009 a — .001 am 2 

47. .001 x 2 ° - .004 y 2b 

48. .003 x 4c - .027 y 2 


286 


ALGEBRA 


EXERCISE 164 

In each example, find the sum, the product, the difference, 
and the quotient of the two fractions: 


i. 5,5 

8 6 

2 x 3 x 
2 ‘ 5 ’ 4 

„ 5 m 3 n 
3 ' ~2~’ ~5~ 

2xy 3 xy 
9*7 
7 ab 3 ab 

6 ' TF’IF 

6 A,J_ 

5 a 8 a 

7. 

4:X 7x 


8 . 


_3_5_ 

11 m 2 ’ 4 m 2 
7 4 

3 xy ’ 9 z 2 
2_ 7_ 
a6 6c 

E, V. 

r s 
c d 
mn m 2 
2 a 3 6 
Zb’ 2 a 
2 x 3 y 
~5y 2 ’ lOz 2 


15. 

16. 

17. 

18. 

19. 

20 . 

21 . 


A, A 

a 2 6* a6 2 
1 2 
z + 2 A - 1 
3 4 

a — b a + b 
2 x 3 x 

x - y x + y 
m — n m + n 
5 ’ 4 

r — x 3 

7 * r + # 
c — d c + c? 
c cf 


Simplify by performing the indicated operations: 

16 m 2 




25. 4 m - 3 - - 


10 


23. a - 4 + 


2 + 11 a 


3 a 


2 y z 


24. x 2 + xy + y 2 - 

x - y 

2 m 


26. m — n — 


27. 3 c + 1 - 


4m + 3 
m 3 — n 3 
m 2 + mn + n 2 
1 - 5c 
4c - 3 


28. 


3 m 


m 2 + m — 6 m 2 + 3 m 


2 m 


1 — m 


2 m + 3 2 


2 m 2 — m — 6 


^ + 2+^ hh 2+^ 
y 2 z 2 j 12/ z 

s >( 6 + 5 #^rM 2 \Fr) 














SUPPLEMENTARY EXERCISES 


287 


32. (m + l - -Ufm-l--') 

\ a — b) \ a + 2 b) 

34. —-- S- (l + _M 

a+ b V a-b) 

36. (a - * (i + ar \ 

36. (— - 2 + H- (- - 

\3 y 2 x) \y x) 

OT 7 fr, „ Q 5 X — 4\ /X — 2 1 — X \ 

37 - \ 2X ~ 3 "332i) ^ (— -47=l) 


38. 

39 

40 

41 


t _j_ a + r 


a + r ’ t — a 
2 _ 

Sx + 2 


(a: 2 — 


(05 - t)(a + r) 
x — l) (a: * 2 


5 x + 4 


\ a: 2 — 3 a: — 4 

m 2 + 6 mn +9 n 2 
m 2 -f- 4 mn + 4 n 2 

2 - 10 a 


— --— \ 

3 a: 4 2 x — 2) 

K l — x ~ ( x2 + 6 x + 8 \ 

x — x 2 ) \ x 2 H- x ) 

m 2 — mn — 6 n 2 > 


m 2 — 0 n 2 

4^ 


\m l — 


42. a - 5 


43. - 3 


3 a 


44. 4 - 


m + n 
3 a 


3 a b 


45. 2 x - 1 - 

46. a — b — 


3 + s 
2 x 3 
a 3 — ft 3 


47 . 


a 2 + ab + b 2 
2 


48. 


49. 


50. 


51. 


52. 


m 2 + mn 

c (Z 


— b n z \ 

- 6 nv 


2c + 2(Z 4c— 4 d 

a 2 — 4 05 + 4 


Or 


a" 


4 a 4 + 2 a 3 + a 2 


2 n 


5 n 


3m—4 4m + 3 
4 05 2 05 


2 — 6 x + 8 * a: 2 — 
e — 4 2 a: — 5 


a: 


4- 


4 a: - 6 12 - 15 x 


3 x — 6 8 — 4 a: 

53. - -W (- + ~) 

\S xJ \s X/ 


12 









































288 


ALGEBRA 


EXERCISE 165 

1. Mv - l)-*(5y + 7)=^ 

2. i(x - 2) -i(x - 4) = f 

3. K4 6 + 1) - i(2 6 + 3) = i(5 6 - 1) 


4. 


6. 


a 


2 a — 3 

1 


- 3 


= 0 


5 a + 1 
2 a — 3 

5 — x _ 
x — 3 3 — x 

^(ii +o=i +|(io-0 

1 V(7w + l)-i( W -9) = l 

i( r + 2 ) —■£(»• — 2 ) = 2 


10. i(t - 1) + 3 = i(t + 14) 

11. |(w> + 12) = %(w - 9) 

3 a + 1 cl -\- 2 _ 


12 . 


6 . 

7. 

8 . 

9. iO + 18) - f 0 - 3) = 4 

15. ^(3 m — 5) — ^(9 w — 7) = f m 

16. i(m + 11) - -i(10 - m) = - 1 


13. 


14. 


3 2 a + 5 

1 . 5 # — 4 1 , 0 CN 

7 3 x + 1 14 V ; 


5t 2 + 4t- 


5,+ 8 -SO 5 '- 13 ) 


17. ±(ar + 18) - f(x - 3) = 4 
lg 4^-36, 11 11 


19. 


20 . 


21 . 


22 . 


23. 


a: 2 — 

9 

+ 3- 

x x + 3 


6 


a — 6 

26 - a 2 


a - 4 


ci + 3 

a 2 — a — 

12 

2/ 

— 

7 

1 


2/ 2 + 3 1/ 

- 28 

4 -y 

.V 

17 


9 

x + 17 


2 # — 

16 

a: 2 — 

to 

1 

00 

2 

5 


10 

7 m + 1 


3 — m 


m + 3 

m 2 — 9 


t 


5 

4 —/ 2 



13 


2* + 12 


24. 


t + 5 
,4+2 

,4 +3 3 — 2,4 2+ 2 + 3v4-9 


* - 2 10 - 3 t - t 2 

A- 1 _ 3+ 2 + 6 





























a i a 


SUPPLEMENTARY EXERCISES 


289 


EXERCISE 166 — Literal Equations 

Solve the following for x : 


, x -\- b a x + a _ Q 

1 * - — - — O CL 


2 . ?- ~ .IJ= a: + 8c -2c 


10 


0 x + ll r 10 r — x 

6 3 

. x , lOt — x 

4. ---h t =--- 


x — m 


= 3 8. -JL. = i 11. 5 - * a 


x - b C 
— b = c 


x — a a 

9. -J_-i 12. 

x — a x 
c b 


10 . 


13. 


4: — X b 
m — x _ c 
n — x d 
c d 


x — a x 


a — x b — x 


„. x — 2 a 4 a 0 

14. —-- -- Z 

5 x x 


16 4 a — x _ j _ 12 a 


a — x 


3 a — x 


15. 


ab 


a 2 + 3 b 2 x 3 x + 4 


17. JiL±A + 3 = X 


2 x — 3b 


2x — 3b 


18 5 x + 4 m _ 11 x — 2 m _ ^ 

2 x Ox 

19 3 x — 5 A 9 x — 7 A 2 

20 . 

21 . 

22 . 


23. 


4 

12 

3 x 

i 

*ss 

1 

a 

10 x + 9 k 

= 51 

5 

10 

6 x 

6 x + 5 c 

X 2c . 

3 x 

2 x 2 — 2 cx 

H 

H. 

1 

% 

i 

x 2 — c 2 

2 x 

4 x + 5 t 

3 t 

1 

1 

a 

CO 

Ox — t 

1 

a 

CO 

9 

2 

1 

3 x — 5 r 

1 

<N 

1 

a 

v — 3 r 

mx —2 m 2 

= mn — nx 

— n 2 

a 2 (x - 1) = 

= b(b — 2 a 

+ ax) 









































290 


ALGEBRA 


EXERCISE 167 


2 . 


I 3 x + ay = 5 
1 2 x — by = 6 
J 4: x + my = 

1 x + py = q 
[ 2 ax -{- y — b 
\ ax — 2 y = c 

TO£ + 717/ = 1 
cx — y = 2 
\ax — by = 3 
} + ay = 1 

f3z , y 


10 . 


21 . 


17. 


18. 


19. 


20 . 


Solve the following systems for x and y: 

6 j cx - dy = a 
' \ x + by = c 
mx ny = m 
x — my = n 
rx — sy — t 
sx + ry = t 
f 3 ic — 4 7/ = .5 
+ 2y = .8 
\5x — Sy = .1 
\ 2x + by =1.9 
f 10 _ 9 = 4 

I x y 

8 _ 15 _ 9 
[x y 2 

[° + i° —i 

I x y 

6 + l-5 = 1 

[x y 
[6 + 4 = 4 
\x y 5 

9 , 5 = 7_ 

[a; y 10 

[2* +- = -ii 
I y 

4x-? = 21 

l y 2 

[- — 1 = 9 
x y 

U+? —1 


+ | = 11 

I a b 

I 5 x _ y _ 13 

[ a b 

(4® _j_ 3^ = _ 1 

| a 6 

5® . V = 7 

[ a 6 

_ 6j/ = _ 

) m n 

j 2 % _7 ?/ _ _ 


10 


23. 


15 


12 . 


13. 


| 4z + 3 y = 6.4 
j 3 a: — .5 7 / = 1.5 
f 5a: - .2 7 / =8.2 
[ 3z-K4t/ = 9.6 
j ax — by = b 
\bx -\- ay — a 
j ax -{-by = c 
[ for — ay = c 
j mx + ny = d 
[ Tiz — my = e 
f a; 4 2 y _ 7 a 


[6_z 2j/ = _ 3 

j c d 

I ^ _ 3 j/ = _ Q 
l c d 


24. 


f3x = 1 
| * r 

I - + ^ = 12 


25. 


26. 


27. 


28. 


29. 


30. 


6 


|i _ 0JL = & 
[3 6 

I 7 14 

H+if- 8 - 

| x y 
I 5_6 = _1 
[x y 2 
f 3 % _ 3 y _ 

| a b 

—+= 2 
[a b 

\ x - c = d - y 

\ d 2 c 

k±i=3 




SUPPLEMENTARY EXERCISES 


291 


EXERCISE 168 


Simplify each radical and combine the results into a single 
term: 


1. 2 V 2 - 3 V 18 + 4 V 50 

2 . 3 V 1 O 8 - ^v / 243 + 2v / 75 

3. 5 VTl + 2 V 44 - V 99 

4. 6 V 1 + |V8 + 7 Vz$- 
6. V48 - 2Vl2 + lOV^ 


6 . 



2 aV 12 


9. V2~a? -Vl8 - iaV 8 

10 . 6 m 4 - V 54 +V 24 & 



7 . 


'a 'x 


2 Vox 3 


8 . Wi + 3 Vi + 3V8 


13. 2(4 — li(4 c)i + 

14. + 2 x^/~x + 

X 


15. fv /24 « 3 6 2 + 6V60 3 - 2 aV 6 ab 2 

16. V(o + £>) 2 f/ + (0 — b) V4 3 / — V 9 a 2 j/ 

17. 4 av / 128 - 3Vl62 a 2 + ^ 


Multiply the following and simplify the result: 

18. (2V5 - 3V2) by (3V5 - 2V2). 

19. (5V6 -V2) by (2V6 +V2). 

20. (2V3 -Ve) by (3V3 + 2V6). 

21. (3V2 -VT6) by (2V2 - 3VT0). 

22. ( 2 V 3 - 3V2) by (2V3 + 3V2). 

23. (2Vx - l) 2 . 24. (aVb + bVa) 2 . 

26. 2 V 3 V 6 - 5V12. 26. V5(V5 - 2)(V5 + 2). 

27. (Va+Va — l)(Va—Va — 1). 









292 


ALGEBRA 


EXERCISE 169 

Miscellaneous Algebraic Expression Drill 

1. A boy bought d dozen oranges at s cents an orange, and 
had c cents left. How much had he at first ? 

2. If the length of a room is L yards, the width W yards, and 
the height H yards, write the expression which represents the 
total surface of: a. the four walls ; b. the ceiling. 

3. If C represents the cost of an article, and r the rate per 
cent of gain, express the selling price in terms of C and r. 

4. The perimeter of a certain square is 2 s inches. Express 
the area of the square. 

6. A man walks m miles in h hours. How far can he walk in 
3 hours ? 

6. If a man can do a piece of work in d days, what part can 
he do in 2 days ? 

7. What is the amount if P dollars is invested at r per cent 
simple interest for 5 years ? 

8. 2 n + 1 is an odd integer. Express the two larger consec¬ 
utive even integers; the two larger consecutive odd integers. 

9. Write in symbols : the square of twice a number, diminished 
by twice the square of the same number. 

10 . If the width of a rectangle is represented by W feet, 
represent the width of a rectangle which is : a. 5 feet narrower; 
b. 5 feet wider; c. 5 times as wide; d. one fifth as wide. 

11. If tennis balls cost r cents a dozen last year, and the price 
has advanced 60^ per dozen, how much will one half dozen cost 
at the present price ? 

12. In how many hours can a boy walk m miles at the rate 
of c miles per hour ? 

13. Sugar is being sold at 9^ per pound. How many pounds 
can be got in exchange for e dozen eggs at y cents per dozen ? 


SUPPLEMENTARY EXERCISES 


293 


14. How many feet are there in y yards, / feet, and i inches ? 

15. If a man’s rate of rowing in still water is s miles per hour, 
and the rate of the current of a river is t miles per hour, what 
is his rate when rowing on that river: a. downstream; b. up¬ 
stream ? 

16. Express the number whose digits in order are a, x, m, and y. 

17. How many hours are there in d days ? 

18. If p pounds of tea cost m dollars, how many cents per 
pound did the tea cost ? 

19. If a merchant paid d dollars for n yards of velvet, what 
was the average price per yard ? 

20. How many pounds are there in z ounces ? 

21. How many ounces are there in p pounds? 

22. If an automobile goes m miles in one hour, what is its rate 
per minute ? 

23. How many yards are there in k inches? 

24. A boy will be k years old 2 years hence. How old was he 
5 years ago ? 

25. If the list price of a certain article is L, and the rate of 
discount is d per cent, what is the selling price ? Represent this 
selling price by S, thus making a formula for the selling price. 

26. What is the total cost of n pounds of corn at c cents per 
pound, m pounds of wheat at w cents per pound, and k pounds 
of barley at b cents per pound ? 

27. What part of a is b ? What per cent of a is b ? 

28. What was the cost per mile for gasoline if an automobile 
consumed N gallons, at p cents per gallon, in going M miles ? 

29. In a class, there are b boys and g girls. What per cent of 
the total membership is the number of boys ? 

30. Express the number whose hundreds’ digit is n, whose 
tens’ digit is 8, and units’ digit is m. 


294 


ALGEBRA 


EXERCISE 170—Interest Problems 

First Degree Equations — One or Two Unknowns 

1. One sum of money is invested at 5%; a second sum is 
invested at 6%. If 3 times the first sum exceeds the second 
sum by $100, and if the total income is $155, find the sums 
invested. 

2. A man has $5000 invested at 4%. How much money 
must he invest at 6% to make the total income equivalent to 
5% on the total amount invested? 

3. A man has $3000 invested, at 3.5%, and $4500 at 4%. 
How much must he invest at 6% to make the total income 
equivalent to 5% on the total sum invested? 

4. A man has $3000 invested, part at 5% and part at 6%. 
His total income per year is $157. How much has he invested 
at each rate ? 

5. The income at 5% on one sum of money exceeds by 
$35.50 the income at 4% on a sum which is $350 less than the 
first. Find the two sums invested. 

6. The income on one sum of money at 4j% and the income 
on a sum $600 greater at 3|% together amount to $421 per year. 
Find the total amount invested. 

Hint. — 4 ^ % = f % = -5^5-. 

7. A man invests a sum of money in 4|% stock and a sum 
$180 greater than the first in 3j% stock. If the incomes from 
the two investments are equal, find the sums invested. 

8. A man has $8000 invested at 5% and $5000 at 7%. 
How much must he invest at 8% to make his income equal to 
6% on his total investment ? 

9. A man has $2500 invested, from which he receives a 
total income of $135. Part of the money is invested at 6% 
and part at 4j%. How much is invested in each way? 


SUPPLEMENTARY EXERCISES 


295 


EXERCISE 171 — Age Problems 

First Degree Equations — One or Two Unknowns 

1. A father is now 9 times as old as his son. In 9 years he 
will be only 3 times as old as his son. What are their present ages ? 

2. The difference between the present ages of a father and 
son is 25 years. In 10 years the father will be twice as old as 
his son. What are their present ages? 

3. A is 5 times as old as B. In 9 years he will be only 3 
times as old as B. What are their ages? 

4. B is twice as old as A. 35 years ago he was 7 times as 
old as A. What are their present ages ? 

5. A is 68 years of age, and B is 11. In how many years 
will A be 4 times as old as B ? 

6. The age of a father is 5 times that of his son; his age 5 
years from now will exceed 3 times his son’s age by 4 years. 
Find their present ages. 

7. A’s age is three eighths of B’s, and 8 years ago it was two 
sevenths of B’s age. Find their ages at present. 

8. Washington was admitted to the Union 18 years before 
Oklahoma, and may therefore be said to be 18 years older than 
Oklahoma as a state. One fourth of Washington’s age in 1911 
exceeded Oklahoma’s age by 1| years. Find the year when 
each was admitted. 

9. In 1912, the “ age ” of Maine was 4 T 2 T times that of 
Wyoming; in 1920, it was 3-J times it. Find when each state 
was admitted to the Union. 

10 . A’s age 11 years from now divided by his age 11 years 
ago is the fraction Find his present age. 

11. A’s age exceeds twice B’s age by 7 years. The sum of 
three fifths of B’s age 2 years ago and of four sevenths of A’s 
age 4 years from now is 26 years. Find their present ages. 


296 


ALGEBRA 


EXERCISE 172 — Distance, Rate, and Time Problems 

1. Suppose that two men, who travel at the rate of 6 miles 
and 10 miles per hour respectively, start from the same place 
in opposite directions. In how many hours will they be 200 
miles apart? 

2 . Suppose that A, traveling 10 miles per hour, leaves a 
place 3 hours before B; suppose that B travels 15 miles per 
hour. In how many hours will B overtake A ? 

3. Two hours after A left, B starts after him in an auto¬ 
mobile at the rate of 27 miles an hour and overtakes him in 
hours. At what rate was A traveling ? 

4. A and B travel toward each other from points separated 
by 250 miles, A at a rate which exceeds B’s rate by 8 miles an 
hour. If they meet in 5 hours, at what rate did each travel ? 

5. Some boys who are boating on a river know that they 
can go with the current 6 miles per hour and can return against 
the current at the rate of 3 miles per hour. How far may they 
go if they have only 3 hours for the trip ? 

6. A man has 11 hours at his disposal. How far may he 
go in a buggy at the rate of 10 miles an hour if he plans to return 
at an average rate of 7 miles per hour ? 

7 . An automobile is traveling at the rate of 25 miles an hour. 
In how many hours will a second automobile overtake the first 
if the second starts 2 hours later than the first, and travels at 
the rate of 35 miles an hour? 

8. An express train, whose rate is 36 miles an hour, starts 
54 minutes after a slow train and overtakes it in 1 hour and 
48 minutes. What is the rate of the slow train? 

9. An automobile party is traveling at the rate of 20 miles 
per hour. At what rate must a second automobile travel in 
order to overtake the first if it starts 2 hours after the first and 
wishes to overtake it in 3 hours ? 


SUPPLEMENTARY EXERCISES 


297 


10. Chicago and Madison, Wisconsin, are about 140 miles 
apart. Suppose that a train starts from each city toward the 
other, one at the rate of 35 miles per hour and the other at the 
rate of 40 miles per hour. How soon will they meet ? 

11. The rate of an express train is three times that of a slow 
train. It covers 180 miles in 8 hours less time than the slow 
train. Find the rate of each train. 

12 . A messenger starts out to deliver a package to a point 
24 miles distant, at the rate of 8 miles per hour. At what rate 
must a second messenger travel to arrive at the same time as 
the first messenger, if he starts 1 hour after him ? 

13. The rate of a passenger train exceeds twice the rate of 
a freight train by 5 miles per hour. It can go 350 miles while 
the freight train goes 150 miles. Find the rate of each train. 

14. Two men, A and B, 57 miles apart, travel towards each 
other, B starting 20 minutes after A. A travels at the rate of 
6 miles an hour and B at the rate of 5 miles an hour. How 
far will each have traveled when they meet ? 

15. The rate of an express train is five thirds of that of a slow 
train. It travels 36 miles in 32 minutes less time than the slow 
train. Find the rate of each train. 

16. The rate of an express train is three halves that of a 
slow train. It covers 270 miles in 3 hours less time than the 
slow train. Find the rate of the trains. 

Quadratic Equations 

17. A man traveled 105 miles. If he had gone 9 miles more 
an hour, he would have performed the journey in lj hours less 
time. Find his rate in miles an hour. 

18. An automobile made a trip of 50 miles, 10 miles within 
city limits and 40 miles outside city limits. Outside city limits, 
the rate was increased 15 miles an hour. If the trip took 2f 
hours, find the rate within and outside city limits. 


298 


ALGEBRA 


EXERCISE 173 — River Problems 

1. Some boys who can row 4 miles an hour in still water 
made a trip on a river whose current is 2 miles an hour. If it 
took them 8 hours for the trip, how far did they go ? 

2. A party took a trip in a motor boat which runs at the 
rate of 15 miles an hour. They took three hours for the round 
trip. What distance did they go, if the rate of the current is 
3 miles an hour ? 

3. Some boys row on a river whose current is known to be 
2 \ miles per hour. They find that it takes them as long to go 
upstream 2 miles as downstream 7 miles. Find their rate. 

4. Some boys are rowing on a river whose current is 5 miles 
per hour in one stretch and 3 miles in another. They find when 
going downstream that they can go 4 miles where the current 
is rapid in the same time that they can go 3 miles where the 
current is slower. Find their rate in still water. 

5. A crew can row 10 miles downstream in 50 minutes, and 
12 miles upstream in an hour and a half. Find the rate in miles 
an hour of the current, and of the crew in still water. 

6. A motor boat which can run at the rate of 15 miles an 
hour in still water went downstream a certain distance in 4 
hours. It took 6 hours for the trip back. What was the dis¬ 
tance and the rate of the current? 

7. A crew is rowing on a stream the rate of whose current 
is known to be 2 miles an hour; they find that it takes them 
one and one third hours to go down, and four hours to come 
back a certain distance. Find the distance and the rate of the 
crew in still water. 

8. A crew can row 8 miles downstream and back again in 
4-f hours; if the rate of the stream is 4 miles an hour, find the 
rate of the crew in still water. 


SUPPLEMENTARY EXERCISES 


299 


EXERCISE 174 — Digit Problems 

1. The tens’ digit of a number exceeds its units’ digit by 
4. If the digits be reversed, the new number is 6 more than 
one half of the old number. Find the number 

2 . The sum of the two digits of a number is 9. If the 
digits be reversed, the quotient of the new number divided by 
the units’ digit of the given number is 13 and the remainder 
is 1. Find the number. 

3. The sum of the two digits of a number is 16; and if 18 
be subtracted from the number, the remainder equals the 
number obtained by reversing the digits. Find the number. 

4. If the digits of a number of two figures be reversed, the 
sum of the resulting number and twice the given number is 
204; and if the given number is divided by the sum of its 
digits, the quotient is 7 and the remainder is 6. Find the 
number. 

5. If a certain number be divided by the sum of its two 
digits, the quotient is 4 and the remainder is 3. If the digits 
be reversed, the sum of the resulting number and 23 is twice 
the given number. Find the given number. 

6. The sum of the two digits of a number is 15. If the 
digits be reversed, the new number exceeds two thirds of the 
original number by 35. What is the number ? 

7. A number consists of three digits of which the hundreds’ 
digit is 2. The sum of the digits is 13. If the digits be reversed, 
the new number exceeds 3 times the original number by 118. 
What is the number ? 

8. A number consists of three digits of which the tens’ digit 
is 5. If the digits be reversed, the sum of the original number 
and the new number is 1110. The number exceeds 23 times 
the sum of its digits by 12. What is the number ? 


300 


ALGEBRA 


EXERCISE 175 — Making Formulas 

1 . a. Express as a formula the number of days, D , in w 
weeks, m months (of 30 days each), and y years (of 360 days 
each). 

6 . Find D when w = 5, m = 7, and y = 3. 

2 . a . Express as a formula the number of cents, C, in n 
nickels, d dimes, q quarters, and D dollars. 

b. Find C, when n = 3, d — 6, q = 4, and D = 5. 

3. a. Express as a formula the number of pints, P, in q 
quarts, n pints, and g gallons. 

b. Find P when n — 3, q = 5, and g — 7. 

4. a. Express as a formula the number of inches, I, in i 
inches, / feet, and y yards. 

b. Find I when i = 7, / = 6, and y = 9. 

5. a. Express as a formula the number of quarts, Q, in 
q quarts, p pecks, and b bushels. 

b. Find Q when q = 3, p = 7, and 6=8. 

6 . Express as a formula the relation between the selling 
price, S, the cost, C, and the loss, L. 

7. Express as a formula the relation between the selling 
price, S, the cost, C, and the gain, if the gain is g% of the cost. 

8 . a. Express as a formula the relation between the selling 
price, S, the cost, C, and the gain, if the gain is g% of the selling 
price. 

6 . Find S if C = $13.75 and g = 15. 

9. a. Express as a formula the total income, T, of a man who 
receives a salary of S dollars per year, and interest at r% on 
investments of P dollars. 

6 . Find T when S = $3750, P = $4500, and r = 6. 


SUPPLEMENTARY EXERCISES 


301 


10 . Express as a formula the total yearly cost, C, of living 
in a home, if the taxes are t dollars per year, the repairs are p 
dollars per year, the insurance is i dollars per year, and interest 
on the cost of the home is r% of V dollars. 

11. A man ran his automobile n miles during the year. He 
spent g dollars for gasoline, o dollars for oil, r dollars for repairs, 
l dollars for license fee, t dollars for tires. He estimated the 
depreciation as d dollars, and charged himself R% of V dollars 
as interest on the present value of the car. Express by a 
formula the cost per mile for running the car. 

12. a. Express as a formula the cost, C, of excavating a cellar 
l feet long, w feet wide, and d feet deep, at n cents per cubic yard. 

b. Find the cost, C, when l = 38, w = 20, d = 7J-, and 
n — 50. 

13. a. Express as a formula the cost, C, of a fence for a lot 
l feet long and w feet wide, if the fence costs m cents per rod. 

b. Find C when l = 345, w = 150, and m = 32. 

14. a. Express as a formula the cost, C, of a cement sidewalk 
l feet long and w feet wide, at n cents per square yard. 

b. Find the cost if l = 800, w — 5, and n = 95. 

15. Express as a formula the cost, C, of plastering a room 
l feet long, w feet wide, and h feet high, at n cents per square 
yard. 

16. a. Express as a formula the length, L, of the two fences 
which inclose a circular track whose inside diameter is d feet, 
and whose outside diameter is D feet. 

b. Find L when d = 410, and D = 440. 

17. a. Express as a formula the average, A, of four numbers 
r, s, t, and w. 

b. Find A when r = 18, s = 26, t = 34, and w = 28. 

c. Find s, if A = 31, r = 28, t = 46, and w = 35. 




302 


ALGEBRA 


EXERCISE 176 


State whether each of the following statements is true or false: 

1. The area (A) of a triangle is A = j bh. 

2. If d is constant in the formula s = then s increases as 


t decreases. 

3. If both numerator and denominator are multiplied by 2, 
the value of the fraction is doubled. 

4. a 2 — 3 a + 9 is a perfect square. 

5. 2 m 2 — 3 x 2 is a binomial. 

6. The sum of two monomials is always a monomial. 

7. The square of a positive number may be less than the 
number. 

8 . A binomial may be a perfect square. 

9. K + - increases as n increases when K is constant and n 

n 

is a positive integer. 

10. The sum of two binomials cannot be zero. 

11. The product of two binomials cannot be zero. 

12. The square of the difference of two numbers equals the 
difference of their squares. 

13. If the sum of two numbers is zero, then their product 
is zero. 

14. The product of two radicals may be rational. 

15. The two roots of an incomplete quadratic equation cannot 
both be negative. 

16. If the width of a rectangle is decreased by n and the 
length is increased by n, then the perimeter is not changed. 

17. Odd powers of negative numbers have the negative sign. 

18. In squaring a binomial by inspection, we square each 
term and add these results. 

19. The difference of the squares of two numbers when 
divided by the difference of the numbers gives their sum as 
quotient. 


SUPPLEMENTARY EXERCISES 


303 


EXERCISE 177 

{Completion Exercises) 

Copy and complete each of the following statements: 

1. An algebraic expression of one term is called a-. 

2. If A = s 2 , and s increases, then A -. 

3. A quantity under a radical sign is called a-. 

4. In A — s 2 , if s is doubled, then A is-. 

5. To double a number is to-it by-. 

6. The sum of the four sides of a rectangle is called its-. 

7. When d represents distance |ind t represents time, then ^ 

represents-. 

8. An equation that has exactly two roots is called a- 

equation. 

9. The numbers that, multiplied, produce a product are 

called-of the product. 

10 . To obtain the square of a number, we multiply the number 

by-. 

11. The product of a number by its reciprocal is-. 

12. An algebraic expression of two terms is called a-. 

13. The product of two monomials is a-. 

14. If a man takes n days to do a job, then he can do-part 

of it in one day. 

15. If the base and the altitude of a rectangle are both in¬ 
creased by x feet, the perimeter will be-by-feet. 

16. In clearing a fractional equation of fractions, we-each 

term by the least common denominator. 

17. If (x — 2){x + 3)= x 2 + (?) — 6, then the missing term 

is --. 

18. In removing the negative parentheses in 2 x —(x — 3), 

we could multiply x - 3 by - instead of subtracting it 

from 2 x. 


304 


ALGEBRA 


EXERCISE 178 (Common Errors) 

State the error made in each of the following problems. When 
possible, give the correct answer. 

1. — 3(x — 1) = — 3 # — 3 


2. x 2 


2 2 • 2 3 = 4 5 
(— 3 x) 2 = — 9 x 
(x + y) 2 = x 2 -f y 2 
x 6 ^ x 3 = x 2 


2 3 = 6 
8 X+ b = b 
X 

2x + y 


10. a — b = b — a 

11 x + 2 = x 

V + 2 y 

12 . V?9 = .3 

13. (2 x) 2 = 2 x 2 

14. = 0 




9. 


= 2 ^ 
x 


16. 


16. 


2 — x 


x - 2 
x — 3 


a — 1 a + 1 
18 y 2 ) = (x - 3y)(x + 6 y). 


17. (x 2 -3 xy 

18 - + - = c L+_^. 
x y x + y 

19. a 2 - 4x 2 ={a - 2)(o + 2). 

20. If - — b = c, then x = a + be. 

a 

21 . If x 2 = a 2 + 6 2 , then x = a + b. 

22. Since 2 4 = 4 2 , then 2 3 should equal 3 2 . 

23. a 2 + b 2 is factorable. 

24. x 3 — xc 2 = x(x 2 — ,c 2 ) = (x — c)(x + c). 

x + 3 = 


26. From x — x ^ = 3 we get 3 x 


26. If ax = b , then x = b — a. 

ab -j- c 7 i 

27. -— = b + c. 


28. If* + * 

2 3 


4, then 3 x + 2 x = 4. 











MISCELLANEOUS REVIEWS 


305 


Review XIV. (Chapters I-IV) 

1. By the formula A = ^ a(b + c), find A when a = 6.4, 
b = 13.5, and c = 7.2. 

2. Add 4 xy — 3 y 2 , 2 x 2 — 2.5 y 2 , and 3 x 2 — 2.4 xy. 

3. Add cx — ay + cfe and dx + 3 ay — ez. 

4. Subtract mz 2 — nx + p from Mx 2 + rx — 5 p. 

5. Simplify 3 x — (5 x — [6 x + 7.5 a — 5] + 4.3 a). 

6. a. A boy saves s cents per week for n weeks, and then 
spends d dollars. How many cents has he left? 

b. Express by an equation the fact that eight times a certain 
number exceeds two times the same number by 43. 

7. There are three consecutive integers whose sum is 144. 
Find them. 

8. Add 3(a + b) — 4(tZ + c) and — 2(a + b) + 9 (d + c). 

9. Solve the equation (3 m — 4) — (2 m + 9) = 15. Check. 

10 . By the formula I = PRT , find how long it will take 
$7500 to earn $1500, if the rate of interest is 5%. 

11. Simplify 32 — \ — z —(—22 — 1)5 

12. Five eighths of a certain man’s income is $2165. What 
is his income ? 

13. Find the numerical value of abc — 2 xy when a = 3, 
b = — 2, c = 4, x = 5, and y = 6. 

14. The length of a certain rectangle is 5 times its width. 
Its perimeter is 360 feet. Find its length and width. 

15. a. How many quarts are there in x bushels and y pecks? 

b. How many inches are there in m feet and n inches ? 

c. How many cents are there in a dollars, b quarters, and 
c dimes ? 

16. Solve the equation 12.5 r — 7 r + 2.25 r — 3.5 r = 26.35. 


306 


ALGEBRA 


Review XV. (Chapters I-V) 

1. From the sum of 3.2 x 2 — 4 xy -f- y 2 and 1.5 a; 2 + 2.5 xy 
— 3 y 2 , subtract 2.7 x 2 — 1.2 xy — 3.4 y 2 . 

2. Multiply x 2 — 2 xy — y? by x 2 + xy — y 2 . Check. 

3. Subtract3 x 2a — 4 x a + 5 from 7 x 2a — 9 x a + 11. 

4. Remove parentheses and combine like terms : 

4.2 m —(3.7 x \2m — 1.2 #])+ -5 m 

6. Find the following products : 

a. (— 3 x b ) • (— 6.2 a; 3 ) b. (2 x n ) • (— 5 x m ) 
c. (-|r) • (+-fz 3a ) 

6. Simplify 2(3 m — 5) — 4(2 m -f- 6) — 5(1.4 — 2.2 m). 

7. One number exceeds another by 7. Six times the larger 
exceeds twice the smaller by 98. Find the numbers. 

8. A is now three times as old as B. A’s age four years from 
now exceeds twice B’s age then by 1 year. Find their ages. 

9. A sum of money amounting to $2.05, consists of nickels 
and dimes. The number of dimes exceeds the number of 
nickels by 7. How many of each coin are there ? 

10. By the formula V = 2 ttR(2 R -j- H) find V when 
R = 7 , H = 13, and tv = 3.14. 

11. What expression must be added to 3 x 2 — xy — 4 y 2 to 
give 6 x 2 + 7 xy — 5 y 2 *t 

12. a. Five times a certain- number exceeds three times the 
same number by 19. Find the number. 

b. If 6 be added to twice a certain number, three times the 
sum is 46 more than twice the remainder when 5 is subtracted 
from the number. Find the number. 

13. A dealer has some candy worth 35^ per pound and some 
worth 60^. He wants to make a mixture of 50 pounds which 
he can sell at 45per pound. How many pounds of each 
kind should he take ? 


MISCELLANEOUS REVIEWS 


307 


Review XVI. (Chapters I-VI) 

1. Simplify 5 x -[.2(x - 3 t/) + .3(a: - 4 y)]. 

2. Solve and check: 10 — 5(3 t — 4) = 6(3 — 2 t). 

3. Multiply 9z+2x 2 -5by4 + 3z 2 -6;z. Check. 

4. Divide 12 a 2 - 28 a + 15 by 6 a - 5. Check. 

5. From the sum 7 xy 2 — 3 y z + 5 x 2 — 2 x 2 y and 2 x 3 
— 4 xy 2 + 2 ?/ 3 , subtract 5 x 2 y — 3 xy 2 — 2 y s . 

6. Multiply x 2n — x n — 1 by x n + 1. 

7. If mg = T + mf, find T when m = 50, g = 32, and / = 28. 

8. Separate 63.5 into two parts such that three times the 
smaller exceeds twice the larger by 9. 

9. A is now 8 years older than B. A’s age 2 years ago was 
twice what B’s age was 4 years ago. Find their ages now. 

10. A grocer wishes to mix enough 90^ tea and 50^ tea to 
make a mixture of 50 pounds of tea which he can sell at 65^ per 
pound. How many pounds of each should he take ? 

11 . A purse contains dimes, quarters, and nickels having a 
total value of $4.15. The number of dimes is three times the 
number of nickels; the number of quarters is 7 more than the 
number of nickels. How many of each coin are there ? 

12. Simplify 3(4 a — 5b)— 2(6 b —3a)— 5(3 b + 2 a). 

13. a. Express by a formula the total weight, W, of an auto 
truck weighing w pounds, loaded with n bags of grain whose 
average weight is E pounds. 

b. If w has a fixed value, how does W change when n in¬ 
creases ? 

14. Divide — 24 # 3 ° + 20 x 2a — 16 x a by — 4 x a . 

15. If S = tLl*' find S when r = 3, 1= - 405, and a = - 5. 

r — 1 

16. One number exceeds another by 5.5. Six times the smaller 
number exceeds three times the larger by 4.5. Find the numbers. 



308 


ALGEBRA 


Review XVII. (Chapters I-VII) 

1. a. Simplify 8 a —[(5 a — 12 6) + (c — 8 a) — (7 b — c)]. 
b. Find the value of the result when a = 2, b = 0, andc = —1. 

2. Solve the equation 2(x — 4) — 3(x — 12) = 33 — 2 x. 

3. a. If to apples cost c cents, how much will x apples cost ? 
b. B walks .5 mile per hour more than A. If A walks n 

miles in 6 hours, how far will B walk in x hours ? 

4. If 25 be added to a certain number, the result is 5 less 
than 7 times the number. Find the number. 

5. a. A merchant sent his errand boy to the bank to get 
change for $20, telling him to get 4 times as many quarters, 
twice as many dimes, 3 times as many nickels, and 5 times as 
many pennies as half-dollars. Can the bank fill the order? 

b. Can they do so, if he changes his request for dimes from 
twice as many to 3 times as many dimes as half-dollars ? How 
many of each coin will he receive ? 

6. The total income from three investments is $493. The 
first sum is invested at 6%; the second sum, which is $500 more 
than the first, is invested at 7%; the third sum, which is $500 
less than twice the first, is invested at 5%. Find each of the 
three sums. 

7. A man’s age 10 years from now will be five times what his 
age was 10 years ago. What is his age now ? 

8. A and B are traveling in opposite directions at the rates 
of 22 miles and 27 miles per hour, respectively. If they started 
from the same place at the same time, in how many hours will 
they be 147 miles apart ? 

9. Solve and check the equation : 

(13 - 4 to)-(3 to + 9) = (2 m - 8)- (6 + 7 to) 

10. By the formula p = 2 a + 2 b, find a when p = 375 and 
b = 45. 


MISCELLANEOUS REVIEWS 


309 


Review XVIII. (Chapters I-VIII) 

1. Simplify 3 x — (5 x — 2[7 x + 9 a — 4] — 3 a). 

2. From the sum of .3 a 2 — 2 ab + b 2 and .5 a 2 — .8 ab 
+ .6 b 2 subtract .6 a 2 — A ab — .3 b 2 . 

3. Divide 20 m 5x — 35 m 3x — 30 m x by — 5 m x . 

4. Solve :2c(c + 7)-(c- 5) 2 = (c + 3)(c - 11). Check. 

5. Factor: 

a . m 2 ^ 3 — 25 £ 3 d. a 2 + 6 2 — 9 c 2 — 2 ab 

4 b. 3 a 3 - 108 a e. a 2 - 4 a + 4 - 9 b 2 

c. 8 a 2 m — 5 am — 3 m /. z 2 — 9 y 2 — 1 + 6 y 

6. A and B started from the same place at the same time 

in opposite directions. A traveled 6 miles per hour more than 
B. At what rates were they traveling if they were 276 miles 
apart at the end of 6 hours ? 

7. Find two consecutive integers such that the sum of their 
squares exceeds twice the square of the smaller by 65. 

8. If x = 3, y = — 2, and z = 2, find the value of 

(x - y) 2 —(a: — y)(y ~ z)+(y - z) 2 . 

9. * Separate 17 into two parts such that the square of the 
larger exceeds five times the square of the smaller by 19. 

10. Which of the numbers 3, — 2, 1, — 1, 0, satisfy the 
equation x 4 — 10 x 2 + 9 = 0 ? 

11 * Solve: A 2 = 15 A + 54. Check. 

12. * Solve for x : x 2 = 11 hx — 30 h 2 . Check. 

13. A is now 32 years old and B is 10. In how many years 
will A’s age be twice B’s age at the same time? 

14. The total income from two investments is $546. One 
sum, invested at 8%, exceeds the other, invested at 6%, by $1400. 
Find the two sums. 


310 


ALGEBRA 


1. Simplify 


Review XIX. (Chapters I-IX) 
a b 2 ab 


b — a 


b 2 


a -f- b 

2. Simplify (6 + + (2 

3. Simplify ~ 3 ^ 4- ^~ 3 

X — S X 2 — xs 

4. a. Reduce to a mixed number : (1) 


( 2 ) 


b. Reduce to a mixed expression : (1) 

h 


12 a - 5 


( 2 ) 


9 y 2 + 5 


4 a 3i/ — 1 

. By the formula V = irh 2 ^r — ^ find V when h = 2.4 and 


6 

r = 8. 

6 . Simplify a 


b. 


x + 3 3 — x x 2 — 9 3 — x 

7. Separate 18 into two parts such that the sum of the 
squares shall be 180. 

8. A man has $1000 invested partly at 5J% and partly at 
7%. His total income is $67. Find the amount invested at 
each rate. 

9. a. If a train goes m miles in x hours, how many feet does 
it go in one minute ? In a minutes ? 

b. If a yards of cloth cost m dollars, what will x yards cost ? 

10. A father is now five times as old as his son. Four years 
ago, he was nine times as old as his son was then. What are 
their ages now ? 

11. Find three consecutive odd integers such that the prod¬ 
uct of the first two exceeds five times the third by 110. 

12. Reduce to lowest terms : ^ 

(x + yr 


13. Simplify 


3a- 4 


4 a — 3 5 a -f- 2 


















MISCELLANEOUS REVIEWS 


311 


Review XX. (Chapters I-X) 

1. Divide 30 x 4 + llz 3 — 5 x — 82 x 2 -j- 3 by 3 x 2 -j- 2 x — 4. 

2. Multiply x 2n — 3 x n — 4 by 2 x n — 1. 

3. Write as a single fraction in its lowest terms, and check: 

/ X , 1 + 3 \_ / X _ 1 + X \ 

vi — x X ) \l — X X ) 

4. * Simplify ---+ ---+ --- 

J x 2 — 3 x 2 b - bx + x 2 4x - 3 - x 2 

5. Factor: 

a. 4 c 3 — m 2 c 2 c. x 2 — a 2 + 14 ab — 49 b 2 

b. 9 r 2 — 42 rs -f- 49 s 2 d. x 2a — x a 

6. a. Solve for h the formula S = 2 tt r(h + r). 

b. How does S change when h increases, if r has a fixed value ? 

7. Solve: fO + 1)- %(x - l)-(x + 2)= —~ - ?• 

8. Solve for x : - -- + - ^ - = 8 a. Check. 

4 5 

9. A is now 20 years older than B. In ten years, B’s age 
will be one half of A’s age at that time. Find their ages. 

10. A man has nickels, dimes, and quarters having a total 
value of $2.90. The number of nickels is four less than the 
number of quarters; the number of dimes is one more than 
-f of the number of quarters. How many of each coin are 
there ? 

11. * Solve for x : 3 x 2 — ax — 4 a 2 = 0. Check. 

12. Twice the width of the Pennsylvania Station in New 
York exceeds its length by 80 feet. Four times the length 
exceeds the perimeter by 700 feet. Find the dimensions. 

13. The width of a certain room is 2 feet more than one half 
its length. If one foot be added to each dimension, the new 
width is three fifths of the new length. Find the original 
dimensions. 












312 


ALGEBRA 


Review XXI. (Chapters I-XI) 

1. Factor: 

a. 4 z 2a - 8z° c. 100 m 4 - 81 t 2 

b. 4 m 2 — mn —3 n 2 d. 6 mx — m 2 — 9 x 2 + 4 

2. Divide 16 x 3a — 14 x 2a — 10 x a by — 2 x a . 

3. Divide x 4 + x 3 — 4 + 6 x — 5 x 2 by x 2 — x + 1. 

4. Simplify (3 + ^|) 

6. a. Solve the formula A = ~(b + c + 4 m) for m. 

6 

b. If b, c, and m have fixed values, how does A change when 
h is doubled ? 

^ _ 0 qX 

6. Reduce to lowest terms : —--• 

6 c 2x - 24 

7. Solve: 3 x + 1 — F(5 x — 7)= \(x — 1)+ x • 

O 

8. Simplify : — \x — 2(y — z)\ — S{x — (2 y — 5 z)\ 

9. * Solve the following equations for x: a. x 2 — 3 x = 40. 

b. 3 x 2 2 cx = 5 c?. c. 6 z 2 = 17 ax + 14 a 2 . 

10. Represent by a bar graph : 


Day 

Mon. 

Tues. 

Wed. 

Thurs. 

Fri. 

Sat. 

No. of stoves sold 

155 

185 

120 

143 

160 

72 


11 . a. A and b are connected by the formula A = f 6. 
Make a table of values of A for b = 0, 5, 10, 15, etc., to 50. 

b. Represent this relation graphically. 

c. If A increases, then b • • *. 

d. If b is trebled, then A • • •. 

12. If a motor boat can travel 12 miles per hour in still water, 
how far downstream can it go and return, if the rate of the 
current is 3 miles per hour, and if the trip down and back must 
be made in 6 hours ? 













MISCELLANEOUS REVIEWS 


313 


Review XXII. (Chapters I-XII) 


1. Factor: 


a. 

b. 

2 . 


m b x 2 — 121 m 5 
5 rx 2 — 405 r 3 

Solve and check: 


c. c 2 - 6 cd + 9 d 2 - 25 e 2 

d. 4 x 2 — a 2 + 6 ab — 9 b 2 

5 x - 3 , 3(x + 2) _ 7(5 - x) 
4 3 6 


3. Simplify (6-^±2)^(2 + -^). 


X 


= 2. 

4. Simplify — 

a -f- b 


a 

a — b 


2 b 2 

b 2 - a 2 


Check: let 


5. Solve for y: 5 y - 7(y + 2) = 8(y + 3)- 6 (y + 1). 

6. Two numbers are in the ratio 2:7. If 2 be added to each 
of them, the ratio of the results is 1: 3. What are the numbers ? 


7. The numerator and denominator of a certain fraction 
are in the ratio 2:5. If 2 be added to the numerator, and 1 be 
added to the denominator, the resulting fraction has the value 
^. Find the original fraction. 

cx + dy = d 
ax — by = a 


8. Solve for x and y : 


• f 2 CC ~ I 2/ 

9. Solve for x and y graphically : { u 

l 3 x — y 


10. Solve for x and y : 


?—i 
y 
6 


b. 


8 


2 
x 

5 _ 6 = _1 
.x y 2 

11. A’s age is three fourths of B’s age; in 12 years, A’s age 
will be six sevenths of B’s age. Find their ages. 


- + - = 3 
x y 

15_4 = 4 

x y 


12. The length of a room exceeds its width by 5 feet. If 3 
feet be added to the length, and 2 feet to the width, the area is 
increased 76 square feet. Find the dimensions. 












314 


ALGEBRA 


1. 


Review XXIII. (Chapters I-XIII) 

a. Write the formula for the total area T of the four 


walls and the ceiling of a room L feet long, W feet wide, and H 
feet high. b. Solve this formula for L in terms of T, W, and H. 

2. Separate $29,500 into three parts which are proportional 
to 2, 3, and 5. 

3. Solve and check: 

i(2x-3)-i(x- !)—(* +2) = 


5x — 1 
12 


4. Simplify each radical and unite the results into a single 


term 


3\f- 


iV2S8 


+Vi(5 y - l) 2 


6. Factor: 

а. 4 c 2 — x 2 — 16 y 2 -f 8 xy b. mx 2a — 2 mx a y b + my 2b 

б. If 3 be added to the numerator, and be subtracted from 
the denominator of a certain fraction, the resulting fraction 
has the value 2; if 3 be subtracted from the numerator, and 4 
be subtracted from the denominator, the resulting fraction has 
the value 1. What is the fraction? 


Solve for x and y : 

2 a _ 3 b 
x y 

3 a 
x 


a. 


1 


— = 6 


2^ _L y = 

a b 


* _ ?j/ = n 
a b 


8. Write as a single fraction in its lowest terms : 

+ dh) ■ o + 1) + ( 2 - 

9. a. The length of a room is L feet and its width is W feet. 
Write the formula for the perimeter, P, in terms of L and W. 

b. Solve the formula, written in part a, for W in terms of L 
and P. Find W, when L = 32^ and P = 110. 

c. If L and W are both doubled, then P is-? 

d. Is P doubled, if L is doubled and W is unchanged ? 










MISCELLANEOUS REVIEWS 


315 


Review XXIV. (Chapters I-XIV) 


1.* Simplify: 


2. Solve and check : 


2 u 8 \ 1 , 6 w + 6 \ 

2 — n — n 2 ' ' \ n 2 - n- 6/ 


5 - *(10 t + 1) = f(( + 3)- 4 * 11 

3 


3. 

4. 

5. 


Solve and check 


+ 1 


9 x 2 


Solve: -J-(4 x — !)—•§■ = 


3x+l 1 - 3 x ' 9 x 2 - 1 

i(*-4)+i(3* + 2). 


Solve graphically: 


x + 2 y = 5 
z - y = 2 


6. a. A*s age now is ti times B’s age. t years ago, A’s age 
was m times B’s age then. What are their ages now ? 

b. Using the results of part a as formulas, find their ages 
when n = 4, m = 7, and t = 5. 


7. A motor boat, capable of traveling 15 miles per hour in 
still water, is starting on a trip on a river whose current is 2 miles 
per hour. How far downstream may the party go, if the trip 
down and back must be made in 5 hours ? 


8. The sum of the digits of a certain number of two digits is 
11. If the digits be reversed, the original number exceeds the 
new number by 27. What is the number ? 

9. One machine can do a certain piece of work in 10 hours, 
and a second in 6 hours. How long will it take the two machines 
together to do the work ? 

10. The length of a certain room is 6 feet more than its width. 
If the width is made 2 feet less and the length is made 3 feet 
more, the area is decreased 10 square feet. What are the 
dimensions ? 

11 . * Solve for x by the factoring method : 

a. 3 x 2 + tx = 10 t 2 b. 7 x 2 — 8 mx = 0 









316 


REGENTS’ EXAMINATIONS 


ELEMENTARY ALGEBRA 

Monday, January 18, 1926 


PART I 


Answer all questions in this part. Each question has 2\ credits assigned 
to it. Each answer must he reduced to its simplest form. 

1. Remove the parenthesis and brackets in the following 
and combine like terms : 7 a —[(5 a — 3)+ 3 a] 

2. What is the value of 15 a 3 — 8 a 2 + 2 when a equals 2 ? 

3. Divide 6 x 3 — 23 x 2 + 29 x — 12 by 2 x — 3 

4. If x diminished by 5 equals 2 x divided by 3, what is the 
value of x ? 


5. Given the formula C = f (F — 32); find F when C is 20. 

6. Factor y 2 — 8 y -f- 16 

7. Factor 25 — z 2 

8. Factor 2 x 2 — x — 6 

9. Express with a positive denominator ——~ 

3 2 

10. Reduce to simplest form-- +- 

a — b a + h 

11. Reduce to simplest form -5- - " x 

F 7c 21c 3 

12. Solve the following equation for x : 

2_x _ 3 x — 5 _ q 
3 2 


13. Solve the following set of equations for x and y : 

x — 4 y = 1 
3 x + 4 y — 19 

14. Find the square root of 33 to the nearest tenth. 







REGENTS’ EXAMINATIONS 


317 


15. 

Simplify V72 


16. 

Simplify 14V^ 


17. 

Simplify 3 V 5 + 2 V 3 + 7 V 5 — 

V 3 

18. 

Solve x 2 — 8 x — 20 = 0 


19. 

If a boat sails v miles in t hours 

, how far will it sail in k 

hours 

? 



20. Given the formula s = ^ (a + 1) ; solve this formula for n. 

A 

PART II 

Answer five questions from this part. Full credit will not be granted 
unless all operations (except mental ones) necessary to find results are 
given; simply indicating the operations is not sufficient. Each answer 
should be reduced to its simplest form. 

21. Find to the nearest tenth the roots of x 2 — 3 x — 5 = 0. 

[ 10 ] 

22. Solve the following set of equations for x and y and check 
one pair of values : 

x - y = 2 

x 2 — 2 y 2 = 7 [8, 2] 

23. A man had invested $1000 in two enterprises. The first 

pays him 6% on his investment and the second 5%. If his annual 
income from both investments is $56, how much did he invest 
in each ? [6, 4] 

24. A travels at the rate of x miles an hour, B at the rate of 

y miles an hour. A can travel in 5 hours as far as B can travel 
in 7 hours. A travels 4 miles an hour faster than B. Find 
x and y. [6, 4] 

25. Two integers are in the ratio 3 : 2. Their product exceeds 
their difference by 22 ; find the two integers. [6, 4] 

26. An estate of $5700 is to be divided among a mother, a 


318 


REGENTS’ EXAMINATIONS 


son and a daughter. The mother is to receive $2500 more than 
the son and the daughter twice as much as the son. Find the 
amount that each will receive. [6, 4] 

27. State whether each of the following statements is true or 
false : [Label each answer with the corresponding letter.] 

а. Every quadratic has two roots. [2] 

б. If 2 is a negative number, x 2 + 5 x + 6 is greater than 

x 2 — 5 x + 6. [2] 

c. The square root of any positive number is less than the 
number. [2] 

d. One of the roots of the equation 3 x 2 + 8 x — 3 = 0 is -J. 

[ 2 ] 

e. The product of two binomials may be a binomial. [2] 

28. If a man swims at the rate of 2 miles an hour, the formula 
d = 2 t represents the relation between the distance ( d ) and 
the time (t). 

a. Construct a taole that will give the values of d correspond¬ 
ing to t = 1, 2, 3, 4, 5, 6. [4] 

h. Plot the graph of this table. [4] 

c. Find and mark by a check (vO a point on the graph that 
shows the time needed to cover 7 miles. [2] 


REGENTS’ EXAMINATIONS 319 

ELEMENTARY ALGEBRA 

Monday, June 14, 1926 


PART I 

Answer all questions in this part. Each question has 2% credits assigned 
to it. Each answer must be reduced to its simplest form. 

1. When r — 5, what is the value of r 2 — 2 r — 15 ? 

2. Simplify 22 — 3(2 x — 4)+(9 x + 3) 

3. Find the quotient when 6 x 3 — 7 x 2 + 14 x — 8 is divided 
by 2 z 2 - x + 4 

4. Factor a 2 — 25 b 2 

5. Factor 4 m 2 — 28 m + 49 

6. Factor 2 x 2 + 9 x — 35 

2 2 

7. Reduce to a single fraction - —- 

3 a — 7 3 a + 2 

8. Express as a single fraction in its lowest terms \ 

7 st 2 ^ 14 t 3 
10 hk 5 h 2 k 3 

9. Solve for n the following equation : = 22 + 

10. What must be the value of c if 8 r 2 + cr — 15 has 4 r — 5 
and 2 r + 3 for its factors ? 

11. If 16 is 4 more than 3 a, what is the value of 2 a — 5 ? 

12. Simplify 

^ 5 

13. Write as a single term 

I 2 V 2 - 2 dV 2 + I 8 V 2 - 5 dV 2 

14. Find the square root of 53 to the nearest tenth. 

15. Solve the following set of equations for x and y : 

4 x — y = 9 

2 a — 3 y = — 23 
Solve the following equation for x : 
x 2 - 10 x - 39 = 0 


16. 






320 


REGENTS’ EXAMINATIONS 


17. Eliminate y from the following set of equations and form 
a quadratic equation in x one of whose members is 0: 

2 x — y = 5 
x 2 — 4 y 2 = 5 

18. Using only one letter, represent two numbers having the 
ratio 2:3. 

19. Solve for b the formula S = --- C and write the answer 
as a single fraction. 

20. What is the value of x at the point where the graph of 
y = 2 x + 4 crosses the z-axis ? 


PART II 

Answer five questions from this part. Full credit will not be granted 
unless all operations (except mental ones) necessary to find results are 
given', simply indicating the operations is not sufficient. Each answer 
should be reduced to its simplest form. 

21. The denominator of a fraction is double its numerator. 
If its numerator is increased by 3 and its denominator decreased 
by 4, the value of the fraction becomes 1. Find the denominator 
of the original fraction. [6, 4] 

22. The sum of the three angles of any triangle is 180 degrees. 
Two of the angles of a certain triangle are in the ratio 6: 5 and 
the third angle is the difference between the other two. Find 
the number of degrees in each angle. [6, 4] 

23. Find the roots of the equation 2 x 2 — 5 x — 4 = 0 correct 
to the nearest tenth. [10] 

24. Two automobiles are 276 miles apart and start at the same 
time to travel toward each other. They travel at rates differing 
by 5 miles an hour. If they meet after 6 hours, find the rate of 
each. [6, 4] 

25. In a barnyard are pigs and hens having in all 50 heads 
and 140 feet. How many animals of each kind are there ? [6,4] 

26. A man travels k miles the first day and increases by 
10 miles each day the distance traveled the preceding day. 



REGENTS’ EXAMINATIONS 


321 


a. How far does he travel the second day? the third day? 

the fourth day ? [3] 

b. Write the formula for the distance d he will travel the nth 
day. [5] 

c. Check by letting k = 50 and n = 2. [2] 

27. Select five of the following statements and state whether 
each of the five selected is true or false : [Label each answer with 
the corresponding letter.] [10] 

a. If both terms of a fraction are positive, decreasing the 
denominator of the fraction decreases the value of the fraction. 

b. If the numerator and the denominator of a fraction are 
equal, the value of the fraction is 0. 

c. The difference between two negative numbers may be a 
positive number. 

d. The equation — — — = 0 has 2 for a root. 

H 16 5 

e. The fourth root of a number is the square of the square 
root of the number. 

/. In a radical expression the index of the root to be extracted 
is always an even number. 

g. A coefficient of a term is a factor of that term. 

28. The following table shows the rates charged by post 
offices for money orders of various amounts : 

Amount of order Cost of sending 

$ 2.50 5 cents 

5.00 7 cents 

10.00 10 cents 

20.00 12 cents 

40.00 15 cents 

a. Construct a broken-line graph showing the relation between 
the cost of sending and the amount of the order sent. [8] 

b. State which is changing more rapidly, the amount of the 
order or the cost of sending it. [2] 


322 


REGENTS’ EXAMINATIONS 


ELEMENTARY ALGEBRA 

Monday, January 17, 1927 


PART I 

Answer all questions in this part. Each question has 2\ credits assigned 
to it. Each answer must be reduced to its simplest form. 

1. If x = 3, find the value of x 2 — 7 x — 9 

2. From 11 x 2 + 7 x — 4 subtract 6 x 2 — 2 x — 4 

3. What number divided by 5 gives a quotient of 13 and a 
remainder of 2 ? 

4. Multiply x 2 -f-2x-j-4 by x — 2 

5. Factor 4 x 2 — 28 xy + 49 y 2 

6 . Factor 3 x 2 + 7 x — 6 

7. Express the product 98 X 102 as the difference between 
two squares. 

8 . Express as a single fraction 

4 x + 3 _ 2 x — 3 
5 6 

9. Express as a single fraction 

a 2 — 9 _ a + 3 
2a + 2 2 

10. Find x when — + — = — 

7 5 5 

11. If x bags of coal weigh y pounds, what will 2 bags weigh ? 

12 . Using only one letter, represent three numbers propor¬ 
tional to 2, 4, and 5, that is, having the ratios 2:4:5 

13. If 5 x — 4 y = 15 and 5 x = 7 y, find y. 

14. Simplify V98 

15. Write as a single term 15 aV 3 — 12 aV 3 + 7 aV3 

16. Solve for P the formula A = P( 1 + rt) 







REGENTS’ EXAMINATIONS 


323 


17. If A travels x miles an hour and B y miles an hour, write 
an equation that expresses the fact that in 5 hours A travels the 
same distance that B travels in 6 hours. 

18. Find the two values of a that satisfy the equation 
a 2 — 6 a = 0. 

19. What value must c have in the equation 2 x + c = 0, if 
3 is a root of the equation, that is, if 3 satisfies the equation ? 

20. If a is greater than b, how does the value of ( a — b) 2 
compare with the value of (b — a) 2 ? 

PART II 

Answer five questions from this part. Full credit will not be granted 
unless all operations (except mental ones ) necessary to find results are 
given; simply indicating the operations is not sufficient. Each answer 
should be reduced to its simplest form. 

21. A rug is 4 feet longer than it is wide and has an area of 
21 square feet. Find the dimensions of the rug. [6, 4] 

22. A farmer sows 13 acres with wheat and oats, obtaining 

a total yield of 496 bushels. If the wheat yields 32 bushels per 
acre and the oats 40 bushels per acre, how many acres of each 
did he plant ? [6, 4] 

23. Find the roots oi x 2 — 7 z + 4 = 0 correct to the nearest 

tenth. [10] 

24. A man invested $4500 in two enterprises, the first paying 

7% and the second 4% annually. If his annual income from both 
is $234, how much has he invested in each enterprise ? [6, 4] 

25. A boy rides away from home in an automobile at the 

rate of 28 miles an hour and walks back at the rate of 4 miles an 
hour. The round trip requires 2 hours. How far does he 
ride ? [6, 4] 

26. Find a fraction whose value is T 9 g and whose denominator 
is 32 less than twice its numerator. [6, 4] 

(Continued on page 324.) 


324 


REGENTS’ EXAMINATIONS 


27. The difference between two positive numbers is 2 and the 
difference between their squares is 8. Find the numbers. [6, 4] 

28. The first column of the following table shows the average 
heights of boys 12 years old whose weights are shown in the 
second column: 


Weight in pounds 


Height in inches 


50 

52 

54 

56 

58 

60 


62 

68 

73 

80 

86 

93 


a. Construct a graph showing the relation of weight to 
height. [8] 

b. Determine from the graph what should be the weight of a 
boy whose height is 55 inches. [2] 


REGENTS’ EXAMINATIONS 


325 


ELEMENTARY ALGEBRA (June 13, 1927) 


PART I 

Answer all questions in this part. Each question has 2 \ credits assigned 
to it. Each answer must he reduced to its simplest form. 

1 . When V = 4 and t = 1 , find the value of (V + t)(V — t ) 

2. Simplify (x — y ) 2 — 2(x 2 + 2 xy — y 2 ) 

3. What is the remainder when 6 a 2 — 7a+12is divided 
by 3 a — 2 ? 

4. In the formula A = —, A = 8 and h — 3. Find b. 

2 

5. Robert was gone m hours on an automobile trip. During 
this time he stopped d hours at the home of a friend. If his speed 
while driving was 30 miles an hour, how many miles did he 
drive ? 


6. Factor 6 a 2 + 12 ab + 8 ac 7. Factor 9 a 2 — 16 

8. Express as a single fraction in its simplest form 

a 2 + 2 a — 3 . 4 a + 12 
b 2 b 2 

9. Express as a single fraction in its simplest form 

3c 2 

c 2 - d 2 d - c 

10. State the operation used in reducing the equation 

—-- = - + 4 to the form 10 x — 15 = 4 x + 80 

4 5 


11. The sum of two numbers is 12 and their difference is 4. 
Find the larger number. 

13. Simplify ^ 


12. Solve for L the formula r = 


14. In an equilateral triangle the area A is expressed by the 
formula, A = where b is the length of one side of the 


triangle. If b = 4, find the area to the nearest tenth. 







326 


REGENTS’ EXAMINATIONS 


15. Write as a single term V24 — 3V6 + 4V54 

16. State whether or not the values, x = 5, y = 10, satisfy 
the following set of equations : 

2 x — y = 0 
4 x -j- 3 y = 60 

17. Solve the following equation for x : 

x 2 — 5 x = — 6 

18. What number added to both members of the equation 
y 2 + 6 y = 15 will make the first member of the equation a 
perfect square ? 

19. A rectangular flower plot is 3 feet longer than it is wide. 
Express by an equation in x the fact that this plot has an area of 
54 square feet. 

20. In the equation a -f- a - = 3, which one of the num¬ 
bers, 4, — 3, 3, — 4, is the root (answer) ? 

PART II 

Answer five questions from this part. Full credit will not be granted 
unless all operations (except mental ones ) necessary to find results are 
given; simply indicating the operations is not sufficient. Each answer 
should be reduced to its simplest form. 

21. Twice a certain number added to another number 
equals 9. Five times the first number multiplied by two times 
the second number equals 40. Find the two numbers and 
check. [6, 3, 1] 

22. Two machinists working at similar machines are paid in 

proportion to the number of parts turned out. A turns out 
24 parts in the same time that B turns out 32. If A earns $36 a 
week, how much does B earn a week ? [6, 4] 

23. In a basketball game f of the points made by the winning 
team was 20 more than the points made by the losing team. 
In all a total of 70 points was made. What was the score ? [6, 4] 



REGENTS’ EXAMINATIONS 


327 


24. At a freshman class candy sale fudge was sold at 60 cents 

a pound and “ Turkish delight ” at 40 cents a pound. In all 
30 pounds of candy were sold. After $1.50 had been deducted 
for expenses, $14.10 was left for the class treasurer. How many 
pounds of each kind of candy were sold ? [6, 4] 

25. In the formula = —-—- 4_ac^ a = 

2 a 

b = — 3, c = — 11. Extract the required square root to the 
nearest tenth and find the value of to the nearest tenth. [10] 

26. A boy skated across a lake with the wind at the rate of 
12 miles an hour. Returning against the wind, he went at the 
rate of 4 miles an hour. He made the entire trip in 1^ hours. 
Find the width of the lake. [6, 4] 

27. State whether each of the following statements is true or 
false: [Write the letters a , b, c, d, e in a column and then write 
the word true or false after each letter.] 

a. If x is an integer, x 2 — 4 x + 4 is always greater than 1. [2] 

h x 2 - 2 xy + y 2 x 2(x + y) = q [ 2 ] 
x + y 2(x — y)(x - y) 

c. 16 + 24 x + 9 x 2 is a perfect square. [2] 

121 

e. The expression 4 a means that a is used 4 times as a 
factor. [2] 

28. A pupil’s average standing in algebra during each of the 
first six weeks of the school year is given in the following table: 


1st week 

86 

4th week 

88 

2d week 

75 

5th week 

90 

3d week 

80 

6th week 

92 


Represent this information by means of a bar graph. [10] 







328 


REGENTS’ EXAMINATIONS 


ELEMENTARY ALGEBRA (January 16, 1928) 


PART I 

Answer all questions in this part. Each question has credits assigned 
to it. Each answer must he reduced to its simplest form. 


1. Find the value of a 3 + 4 b 2 — 7 c + 4 when a = 1, 
h = 2, c = 3 

2. Divide 6a: 3 + 11 a; 2 — 1 by 3 a: + 1 

3. Find the dividend when the quotient is x — 1, the re¬ 
mainder 2, and the divisor 3 x — 1 

4. A ship sails r miles the first day, s miles the second day, 
and t miles the third day. Express the average daily rate. 

5. Factor 2 x 2 — 18 

6. Factor 16 a 2 — 2.4 a + .09 

7. Reduce to lowest terms 

a 2 — 6 a 


a 2 — 7 a + 6 

8. If y and x are positive and y = -, does y become larger 
or smaller as x becomes larger ? 

9. Divide (l + —-—by ——— and express the resulting 

v x — 1 / x — 1 

fraction in its lowest terms. 

10. Solve the following equation for v: 

— — 6 = -- + 2 
3 9 


11. Find the square root of 68.89 

12. Simplify 14V^ 

13. Simplify each term of the follow ing e xpression and com¬ 
bine the results into a single term : hV 27 a — 6 V 3 ah 2 

14. Find the two roots of the equation x 2 — 7 x = — 12 

15. Solve for h the formula k(h — 1)= R 

16. If t tons of coal cost s dollars, indicate the cost of m tons. 

17. If 19 is subtracted from 3 times a certain number, the 
remainder is 110. Find the number. 





REGENTS’ EXAMINATIONS 


329 


18. Indicate which one of the following, 1, 0, -- a 

/ z ,\2 a b a — b 

is the correct answer for ■ ~ —— X --- 

4 6 a(a? - 6 2 ) 

19. If X and Y are positive, indicate which is the greater, 
a or 6: 


а. The square of the sum of X and Y. 

б . The sum of the squares of X and F. 

20. In the following table for a graph determine the value of 


y corresponding 

x • 

to the 

1 

value 

2 

x = 4 

3 

4 


y 

2 

4 

6 

? 

. . . 


PART II 

Answer five questions from this part. Full credit will not be granted 
unless all operations (except mental ones) necessary to find results are 
given; simply indicating the operations is not sufficient. Each answer 
should be reduced to its simplest form. \ 

21. Two motor boats start at the same time and place and 
travel in opposite directions. The ratio of their rates is 2:3. 
In 5 hours they are 100 miles apart. Find the rate of each. [6, 4] 

22. A man invests f of his property at 4%, at 5%, and the 
remainder at 3%. If his resulting annual income is $610, find 
the value of his property. [6, 4] 

23. In a mixture of water and turpentine containing 21 

ounces, there are 7 ounces of turpentine. How many ounces 
of turpentine must be added to make a new mixture that 
shall be 75% turpentine ? [7, 3] 

24. Solve for x and y and check your answers : 

x — y = 5 

xy = 24 [8, 2] 

25. Copy and complete each of the following statements : 

a. x 2 — . . . + 4 is a perfect square. [2] 












330 


REGENTS’ EXAMINATIONS 


b. If x is a positive fraction less than 1, then x 2 is . . . than 

[2] 

c. One root of x 2 — 4 x = 8, correct to the nearest tenth, 

is . . . [2] 

d. In s hours an airplane going at the rate of r miles per hour 
can fly . . . miles. [2] 

e. A quadratic equation in one unknown is an equation of the 
. . . degree and has . . . roots. [2] 

26. State whether each of the following statements is true or 
false: [Write the letters a, b, c, d, e in a column and then write 
the word true or false after each letter.] 

a. (-V2) 2 = 4 [2] 

b. To solve the equation x 2 + px = q by “ completing the 
square ” add p 2 to both members of the equation. [2] 

c. One root of the equation x 2 — 3 x — 10 = 0 is — 2. [2] 

d. If the base of a rectangle is doubled, the area of the rec¬ 
tangle is doubled. [2] 



27. A rectangle is m feet longer than it is wide. Its perimeter 
is p feet. Find its length and width in terms of m and p. [10] 

28. The brake-testing table given below shows the distances 
in which automobiles traveling at different rates should be able 
to stop. 


Speed of car 
15 miles an hour 
20 miles an hour 
25 miles an hour 
30 miles an hour 
35 miles an hour 


Distance in which it should stop 
21 feet 
37 feet 
58 feet 
83 feet 
114 feet 


a. Construct a curve (graph) representing this information. [8] 

b. From the graph determine the speed of a car that stops in 
70 feet. [2] 




ALGEBRA 


331 


Table of Powers and Roots 


No. 

Sqs. 

Sq. 

Roots 

Cubes 

Cube 

Roots 

No. 

Sqs. 

Sq. 

Roots 

Cubes 

Cube 

Roots 

1 

1 

1.000 

1 

1.000 

51 

2,601 

7.141 

132,651 

3.708 

2 

4 

1.414 

8 

1.259 

52 

2,704 

7.211 

140,608 

3.732 

3 

9 

1.732 

27 

1.442 

53 

2,809 

7.280 

148,877 

3.756 

4 

16 

2.000 

64 

1.587 

54 

2,916 

7.348 

157,464 

3.779 

5 

25 

2.236 

125 

1.709 

55 

3,025 

7.416 

166,375 

3.802 

6 

36 

2.449 

216 

1.817 

56 

3,136 

7.483 

175,616 

3.825 

7 

49 

2.645 

343 

1.912 

57 

3,249 

7.549 

185,193 

3.848 

8 

64 

2.828 

512 

2.000 

58 

3,364 

7.615 

195,112 

3.870 

9 

81 

3.000 

729 

2.080 

59 

3,481 

7.681 

205,379 

3.893 

10 

100 

3.162 

1,000 

2.154 

60 

3,600 

7.745 

216,000 

3.914 

11 

121 

3.316 

1,331 

2.223 

61 

3,721 

7.810 

226,981 

3.936 

12 

144 

3.464 

1,728 

2.289 

62 

3,844 

7.874 

238,328 

3.957 

13 

169 

3.605 

2,197 

2.351 

63 

3,969 

7.937 

250,047 

3.979 

14 

196 

3.741 

2,744 

2.410 

64 

4,096 

8.000 

262,144 

4.000 

15 

225 

3.872 

3,375 

2.466 

65 

4,225 

8.062 

274,625 

4.020 

16 

256 

4.000 

4,096 

2.519 

66 

4,356 

8.124 

287,496 

4.041 

17 

289 

4.123 

4,913 

2.571 

67 

4,489 

8.185 

300,763 

4.061 

18 

324 

4.242 

5,832 

2.620 

68 

4,624 

8.246 

314,432 

4.081 

19 

361 

4.358 

6,859 

2.668 

69 

4,761 

8.306 

328,509 

4.101 

20 

400 

4.472 

8,000 

2.714 

70 

4,900 

8.366 

343,000 

4.121 

21 

441 

4.582 

9,261 

2.758 

71 

5,041 

8.426 

357,911 

4.140 

22 

484 

4.690 

10,648 

2.802 

72 

5,184 

8.485 

373,248 

4.160 

23 

529 

4.795 

12,167 

2.843 

73 

5,329 

8.544 

389,017 

4.179 

24 

576 

4.898 

13,824 

2.884 

74 

5,476 

8.602 

405,224 

4.198 

25 

625 

5.000 

15,625 

2.924 

75 

5,625 

8.660 

421,875 

4.217 

26 

676 

5.099 

17,576 

2.962 

76 

5,776 

8.717 

438,976 

4.235 

27 

729 

5.196 

19,683 

3.000 

77 

5,929 

8.774 

456,533 

4.254 

28 

784 

5.291 

21,952 

3.036 

78 

6,084 

8.831 

474,552 

4.272 

29 

841 

5.385 

24,389 

3.072 

79 

6,241 

8.888 

493,039 

4.290 

30 

900 

5.477 

27,000 

3.107 

80 

6,400 

8.944 

512,000 

4.308 

31 

961 

5.567 

29,791 

3.141 

81 

6,561 

9.000 

531,441 

4.326 

32 

1,024 

5.656 

32,768 

3.174 

82 

6,724 

9.055 

551,368 

4.344 

33 

1,089 

5.744 

35,937 

3.207 

83 

6,889 

9.110 

571,787 

4.362 

34 

1,156 

5.830 

39,304 

3.239 

84 

7,056 

9.165 

592,704 

4.379 

35 

1,225 

5.916 

42,875 

3.271 

85 

7,225 

9.219 

614,125 

4.396 

36 

1,296 

6.000 

46,656 

3.301 

86 

7,396 

9.273 

636,056 

4.414 

37 

1,369 

6.082 

50,653 

3.332 

87 

7,569 

9.327 

658,503 

4.431 

38 

1,444 

6.164 

54,872 

3.361 

88 

7,744 

9.380 

681,472 

4.447 

39 

1,521 

6.245 

59,319 

3.391 

89 

7,921 

9.433 

704,969 

4.464 

40 

1,600 

6.324 

64,000 

3.419 

90 

8,100 

9.486 

729,000 

4.481 

41 

1,681 

6.403 

68,921 

3.448 

91 

8,281 

9.539 

753,571 

4.497 

42 

1,764 

6.480 

74,088 

3.476 

92 

8,464 

9.591 

778,688 

4.514 

43 

1,849 

6.557 

79,507 

3.503 

93 

8,649 

9.643 

804,357 

4.530 

44 

1,936 

6.633 

85,184 

3.530 

94 

8,836 

9.695 

830,584 

4.546 

45 

2,025 

6.708 

91,125 

3.556 

95 

9,025 

9.746 

857,375 

4.562 

46 

2,116 

6.782 

97,336 

3.583 

96 

9,216 

9.797 

884,736 

4.578 

47 

2,209 

6.855 

103,823 

3.608 

97 

9,409 

9.848 

912,673 

4.594 

48 

2,304 

6.928 

110,592 

3.634 

98 

9,604 

9.899 

941,192 

4.610 

49 

2,401 

7.000 

117,649 

3.659 

99 

9,801 

9.949 

970,299 

4.626 

50 

2,500 

7.071 

125,000 

3.684 

100 

10,000 

10.000 

1 ,000,000 

4.641 



















INDEX 


Abscissa, 194. 

Absolute value, 35. 

Addend, 26. 

Algebraic expression, 23; value 
of an, 23. 

Angle, of elevation, 262; of de¬ 
pression, 262. 

Antecedent, 183. 

Ascending powers, 45. 

Axis, horizontal, 194; vertical, 194. 

Bar graph, 6, 188. 

Base, 30. 

Binomial, 45; square of a, 124. 
Braces, 30. 

Brackets, 30. 

Cancellation, in an equation, 96; 

in a fraction, 140. 

Changing signs, in an equation, 98; 

in a fraction, 164. 

Clearing of fractions, 169. 
Coefficient, 41; numerical, 41. 
Common factor, 26. 

Complete factoring, 116. 
Conditional equation, 95. 
Consequent, 183. 

Coordinates, 194. 

Cosine of an angle, 266. 

Denominator, 139. 

Dependent equations, 222. 
Descending powers, 45. 

Difference, 50. 

Difference of, 112. 

Digit, 216. 

Dividend, 86. 

Divisor, 86. 


Elimination, by addition or sub¬ 
traction, 206; by substitution, 
208. 

Equation, 17; cancelling terms 
in an, 96; changing signs in an, 
98; complete quadratic, 245; 
conditional, 95; identical, 95; 
linear, 201; of first degree, 95; 
incomplete quadratic, 242; 
quadratic, 134; simple, 95; 
solving an, 17; transposition 
in an, 96. 

Equations, dependent, 222; incon¬ 
sistent, 222; independent, 205; 
simultaneous, 205; system of, 
205. 

Exponent, 30; zero, 88. 

Exponents, law of division of, 88; 
law of multiplication of, 72. 

Expression, algebraic, 23; integral, 
158; mixed, 158; surd, 235. 

Extremes, 184. 

Factor, 1, 26; common, 26; to, 
107. 

Factors, prime, 116. 

Formula, 3, 30. 

Fractions, 139; clearing of, 169; 
equivalent, 150. 

Fundamental operations, 23. 

Graph, bar, 6; broken line, 10; 
smooth line, 12; circular, 190. 

Graph of an equation with two 
variables, 196, 202. 

Graphical representation, 188. 

Grouping, symbols of, 30; factor¬ 
ing by, 128. 


332 



INDEX 


333 


Horizontal axis, 194. 

Identity, 95. 

Incomplete quadratic, 242. 

Inconsistent equations, 222. 

Independent equations, 205. 

Left side of an equation, 17. 

Like terms, 26. 

Linear equation, 201. 

Literal, equation, 181; exponent, 
30. 

Literal number, 1. 

Lowest common, multiple, 148; 
denominator, 150. 

Lowest terms, 139. 

Means, 184. 

Members or sides of an equation, 
17. 

Minuend, 50. 

Monomial, 41. 

Monomials, addition of, 42; divi¬ 
sion of, 88; multiplication of, 
73; square of, 107; square root 
of, 108. 

Negative numbers, 35; addition 
of, 36; division of, 87; multi¬ 
plication of, 38; subtraction 
of, 51. 

Negative term, 41. 

Number, literal, 1; negative, 35; 
positive, 35; prime, 107; un¬ 
known, 17. 

Numerator, 139. 

Numerical, coefficient, 23; value, 
23. 

Opposite quantities, 33. 

Ordinate, 194. 

Origin, 194. 

Parallelogram, 4. 

Parentheses, 30; inclosing terms 
in, 64; removing, 61. 


Polynomial, 45; arranging an, 45. 

Polynomials, addition of, 45 ; divi¬ 
sion of, 92; factoring, 109; 
factoring by grouping, 128; 
multiplication of, 76; square 
root of, 227; subtraction of, 
52. 

Positive number, 35; term, 41. 

Power, 30. 

Powers, ascending, 45; descending, 
45. 

Prime number, 107. 

Principal root, 224. 

Product, 1. 

Proportion, 184. 

Quadratic equation, 134; com¬ 
plete, 245; having two un¬ 
knowns, 254; incomplete, 242; 
solution of, by completing the 
square, 246; by factoring, 134; 
by formula, 250. 

Quadratic surd, 233. 

Quotient, 86. 

Radical, 233. 

Ratio, 183. 

Rationalizing denominator, 237. 

Reciprocal, 219. 

Remainder, 50. 

Removing parentheses, 61. 

Right member, or side, 17. 

Right triangle, 244. 

Root, of an equation, 17; princi¬ 
pal, 224; square, 224. 

Satisfy an equation, 17. 

Signs, change of in an equation, 
98; law of, in addition, 36; 
in division, 87; in multiplica¬ 
tion, 38; of a fraction, 164. 

Simultaneous equations, 205. 

Sine of an angle, 264. 

Solution of simultaneous equa¬ 
tions, 205. 



334 


INDEX 


Square root, approximate, 230; 
of a monomial, 108; of a num¬ 
ber, 228; of a polynomial, 227. 

Subtraction, 50. 

Subtrahend, 50. 

Sum, 26. 

Surd, quadratic, 233; addition of, 
235; division of, 236; multipli¬ 
cation of, 236. 

Symbols of grouping, 30. 

System of equations, 205. 

Table of square roots, 331. 

Tangent of an angle, 259. 


Term, 41; negative, 41; positive, 
41. 

Terms, like, 41; unlike, 41. 
Transposition, 96. 

Trinomial, 45. 

Unknown number, 17. 

Unlike terms, 41. 

Variables, 196; related, 272. 
Variation, 272. 

Varies, directly, 273; inversely, 
276; jointly, 275. 

Vertical axis, 194. 

Vinculum, 61. 






0 






















